Functions algebra

Algebra Functions

You are now deeper in your Algebra journey and you've just been introduced to this term called a "function". What in the world is a function?

Although it may seem at first like a function is some foreign creature in Algebra land, a function is really just an equation with a fancy name and fancy notation.

If you are nervous, Algebra Class offers many lessons on understanding functions. Click here to view all function lessons.

Ok, so getting down to it, let's answer that question: "What is a function?"

A function is a relationship between two variables. The first variable determines the value of the second variable. The value of the first variable corresponds to one and only one value for the second variable.


Yes, I know that these formal definitions only make it more confusing. Let's take a look at this another way.

I always go back to my elementary years when we learned about functions - but never called them functions. We had what was known as an "in and out box". Some teachers now call it a "Function Box" and this is why:

Here's a picture of an algebra function box. Imagine the equation being the center of the function box. You put a number in, the function box performs the calculation and out pops the answer.

algebra function machine

Let's take a look at an example with an actual equation. Here we have the equation:
y = 2x+1 in the algebra function box. (Notice how our equation has 2 variables (x and y)

When we input 3, the function box then substitutes 3 for x and calculates the answer to be 7. (2*3 +1 = 7). This means that the variable y = 7.

algebra function example

The equation y = 2x+1 is a function because every time that you substitute 3 for x, you will get an answer of 7. When x = 3, y = 7 every time. No other number will correspond with 3, when using this equation.

If you input another number such as 5, you will get a different output. When you input 5, you should get 11 because (2*5+1 = 1), so when x = 5, y = 11. No other number can correspond with 5, when substituting into this equation. Therefore, this equation can be labeled a function.

It seems pretty easy, right? It seems like all equations would be considered functions. Take a look at an example that is not considered a function.

Let's take the equation: x = y2.

When we input 4 for x, we must take the square root of both sides in order to solve for y. We end up with y = 2 or -2.

function examples

We cannot say that the equation x = y2 represents a function because when we input 4 for x, we get two different answers for y (2 and -2). Therefore, this does not satisfy the definition for a function: "the value of the first variable corresponds to one and only one value for the second value". We have more than one value for y.

Hopefully with these two examples, you now understand the difference between an equation that represents a function and an equation that does not represent a function. You will find more examples as you study the lessons in this chapter.

So, what kinds of functions will you study? In Algebra 1, we will study linear functions (much like linear equations) and quadratic functions. As you progress into Algebra 2, you will be studying exponential functions.

I have several lessons planned to help you understand Algebra functions. Take a look.


Below is the table of contents for the Functions Unit. Click on the lesson that interests you, or follow them in order for a complete study of functions in Algebra 1.

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  3. Functions

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Functions and Their Notation

A function maps a set of inputs onto a set of permissible outputs. Each input corresponds with one and only one output

Learning Objectives

Connect the notation of functions to the notation of equations and understand the criteria for a valid function

Key Takeaways

Key Points

  • Functions are a relation between a set of inputs and a set of outputs with the property that each input maps to exactly one output.
  • Typically functions are named with a single letter such as f.
  • Functions can be thought of as a machine in a box that is open on two ends. You put something into one end of the box, it somehow gets changed inside of the box, and then the result pops out the other end.
  • All functions are relations, but not all relations are functions.

Key Terms

  • output: The output is the result or answer from a function.
  • relation: A relation is a connection between numbers in one set and numbers in another.
  • function: A function is a relation in which each element of the input is associated with exactly one element of the output.

Functions

In mathematics, a function is a relation between a set of inputs and a set of permissible outputs. Functions have the property that each input is related to exactly one output. For example, in the function [latex]f(x)=x^2[/latex] any input for [latex]x[/latex] will give one output only.

Functions are typically named with a single letter, like [latex]f[/latex]. [latex]f(x)[/latex] is read ” [latex]f[/latex] of [latex]x[/latex]“, and represents the output of the function [latex]f[/latex] corresponding to an input [latex]x[/latex].
The input variable (s) are sometimes referred to as the argument(s) of the function. Consider the following example:

Let [latex]f(x)=x^2[/latex]

Let [latex]x=-3[/latex], then:

[latex]\displaystyle f(-3)=(-3)^2 [/latex]

[latex]\displaystyle =9[/latex]

In the example above, the argument is [latex]x=-3[/latex] and the output is [latex]9[/latex]. We write the function as:[latex]f(-3)=9[/latex].

In the case of a function with just one input variable, the input and output of the function can be expressed as an ordered pair. The order is such that the first element is the argument and the second is the output. In the example above, [latex]f(x)=x^2[/latex], we have the ordered pair [latex](−3, 9)[/latex]. If both the input and output are real numbers then the ordered pair can be viewed as the Cartesian coordinates of a point on the graph of the function.

Another commonly used notation for a function is [latex]f:X\rightarrow Y[/latex], which reads as saying that [latex]f[/latex] is a function that maps values from the set [latex]X[/latex] onto values of the set [latex]Y[/latex].

Functions as a Machine

Functions are often described as a machine in a box that is open on two ends. You put something into one end of the box, it gets changed inside of the box, and then the result pops out the other end. The function is the machine inside the box and it’s defined by what it does to whatever you put into it.

image

Function Machine: A function [latex]f[/latex] takes an input [latex]x[/latex] and returns an output [latex]f(x)[/latex]. One metaphor describes the function as a “machine”, that for each input returns a corresponding output.

Let’s say the machine has a blade that slices whatever you put in into two and sends one half of that object out the other end. If you put in a banana you would get back half a banana. If you put in an apple you would get back half an apple.

image

Fruit Halving Function: This shows a function that takes a fruit as input and releases half the fruit as output.

Let’s define the function to take what you put into it and cut it in half. That is, the function divides the input by two. If you put in [latex]2[/latex] you would get back [latex]1[/latex]. If you put in [latex]57[/latex] you would get back [latex]28.5[/latex]. The function machine allows us to alter expressions. In this example, the function would be written as:

[latex]\displaystyle f(x)=\frac{1}{2}x[/latex].

Functions As a Relation

Functions can also be thought of as a subset of relations. A relation is a connection between values in one set and values in another. In other words, each number you put in is associated with each number you get out. In a function every input number is associated with exactly one output number In a relation an input number may be associated with multiple or no output numbers. This is an important fact about functions that cannot be stressed enough: every possible input to the function must have one and only one output. All functions are relations, but not all relations are functions.

Graphical Representations of Functions

Graphs provide a visual representation of functions, showing the relationship between the input values and output values.

Learning Objectives

Describe the relationship between graphs of equations and graphs of functions

Key Takeaways

Key Points

  • Functions have an independent variable and a dependent variable. Typically [latex]x[/latex] is the independent variable and [latex]y[/latex] the dependent variable.
  • As you choose any valid value for the independent variable, the dependent variable is determined by the function.
  • To graph a function, choose some values for the independent variable, [latex]x[/latex], plug them into the function to get a set of ordered pairs [latex](x,f(x))[/latex], and plot these on the graph. Then connect the points to best match how the points are arranged on the graph. Make sure you have enough points.

Key Terms

  • dependent variable: The dependant variable in an equation or function is that whose value depends on one or more independant variables in the equation or function.
  • independent variable: An independent variable in an equation or function is one whose value is not dependent on any other in the equation or function.
  • graph: A diagram displaying data; in particular one showing the relationship between two or more quantities, measurements, or numbers.

Independent and Dependent Variables in Function Notation

Functions have an independent variable and a dependent variable. When we look at a function such as [latex]f(x)=\frac{1}{2}x[/latex], we call the variable that we are changing, in this case [latex]x[/latex], the independent variable. We assign the value of the function to a variable, in this case [latex]y[/latex], that we call the dependent variable.  Function notation, [latex]f(x)[/latex] is read as “[latex]f[/latex] of [latex]x[/latex]” which means “the value of the function at [latex]x[/latex].”  Since the output, or dependent variable is [latex]y[/latex], for function notation often times [latex]f(x)[/latex] is thought of as [latex]y[/latex].  The ordered pairs normally stated in linear equations as [latex](x,y)[/latex], in function notation are now written as [latex](x,f(x))[/latex].

We say that [latex]x[/latex] is independent because we can pick any value for which the function is defined, in this case the set of real numbers [latex]\mathbb{R}[/latex], as inputs into the function. We say the result is assigned to the dependent variable since it depends on what value we placed into the function.

Graphing Functions

Example 1:  Let’s start with a simple linear function:

[latex]\displaystyle f(x)=5-\frac{5}{2}x[/latex].

Start by graphing as if [latex]f(x)[/latex] is a linear equation:

[latex]\displaystyle y=5-\frac{5}{2}x[/latex]

We choose a few values for the independent variable, [latex]x[/latex].  Let’s choose a negative value, zero, and a positive value:

[latex]\displaystyle x=-2, 0, 2 [/latex].

Next, substitute these values into the function for [latex]x[/latex], and solve for [latex]f(x)[/latex] (which means the same as the dependent variable [latex]y[/latex]):  we get the ordered pairs:

[latex]\displaystyle (-2,10), (0,5), (2,0)[/latex]

This function is that of a line, since the highest exponent in the function is a [latex]1[/latex], so simply connect the three points. Extend them in either direction past the points to infinity, and we have our graph.

The line has negative slope and has y-intercept 5 and x=intercept 2.

Line graph: This is the graph of the function [latex]f(x)=5-\frac{5}{2}x[/latex].  The function is linear, since the highest degree in the function is a [latex]1[/latex].  Only two points are required to graph a linear function.

Example 2:  Graph the function:

[latex]\displaystyle f(x)=x^{3}-9x[/latex].

Start by choosing values for the independent variable, [latex]x[/latex]. This function is more complicated than that of a line so we’ll need to choose more points. Let us choose:

[latex]\displaystyle x=\{0, \pm1, \pm2, \pm3, \pm4\}[/latex].

Next, plug these values into the function, [latex]f(x)=x^{3}-9x[/latex], to get a set of ordered pairs, in this case we get the set of ordered pairs:

[latex]\displaystyle \{(-4, -28),(-3,0)(-2,10),(-1,8),(0,0),(1,-8),(2,-10),(3,0),(4,28)\}[/latex].

Next place these points on the graph, and connect them as best as possible with a curve. The graph for this function is below.

The cubic polynomial is negative for x-values less than negative 3, then takes the shape of a hill from x = -3 to x = 0 with a peak of around 10 between x = -2 and x = -1. From x = 0 to x = 3 it has a valley shape with the lowest point of about -10 between x = 1 and x = 2, crossing the x-axis at x=3, then continuing up infinitely for x-values greater than 3.

Cubic graph: Graph of the cubic function [latex]f(x)=x^{3}-9x[/latex].  The degree of the function is 3, therefore it is a cubic function.

The Vertical Line Test

The vertical line test is used to determine whether a curve on an [latex]xy[/latex]-plane is a function

Learning Objectives

Explain why the vertical line test shows, graphically, whether or not a curve is a function

Key Takeaways

Key Points

  • A function can only have one output, [latex]y[/latex], for each unique input, [latex]x[/latex]. If any [latex]x[/latex]-value in a curve is associated with more than one [latex]y[/latex]-value, then the curve does not represent a function.
  • If a vertical line intersects a curve on an [latex]xy[/latex]-plane more than once, then for one value of x the curve has more than one value of y, and the curve does not represent a function.

Key Terms

  • function: A relation in which each element of the input is associated with exactly one element of the output.
  • vertical line test: A visual test that determines whether a curve is a function or not by examining the number of [latex]y[/latex]-values associated with each [latex]x[/latex]-value that lies on the curve.

In mathematics, the vertical line test is a visual way to determine if a curve is a graph of a function, or not. Recall that a function can only have one output, [latex]y[/latex], for each unique input, [latex]x[/latex]. If any [latex]x[/latex]-value in a curve is associated with more than one [latex]y[/latex]-value, then the curve does not represent a function.

If a vertical line intersects a curve on an [latex]xy[/latex]-plane more than once, then for one value of x the curve has more than one value of y, and the curve does not represent a function. If all vertical lines
intersect a curve at most once then the curve represents a function.

The top graph, a curved shape which, from left to right, has a peak, twists in as it descends, then a trough, is not a function. Where it twists, a vertical line is shown which crosses the curve at three red points. The bottom graph is a straight line with negative slope, passes the vertical line test, and is a function.

Vertical Line Test: Note that in the top graph, a single vertical line drawn where the red dots are plotted would intersect the curve 3 times. Thus, it fails the vertical line test and does not represent a function. Any vertical line in the bottom graph passes through only once and hence passes the vertical line test, and thus represents a function.

To use the vertical line test, take a ruler or other straight edge and draw a line parallel to the [latex]y[/latex]-axis for any chosen value of [latex]x[/latex]. If the vertical line you drew intersects the graph more than once for any value of [latex]x[/latex] then the graph is not the graph of a function. If, alternatively, a vertical line intersects the graph no more than once, no matter where the vertical line is placed, then the graph is the graph of a function. For example, a curve which is any straight line other than a vertical line will be the graph of a function.

Example

Refer to the three graphs below, [latex](a)[/latex], [latex](b)[/latex], and [latex](c)[/latex]. Apply the vertical line test to determine which graphs represent functions.

(a) is a cubic polynomial, going up from negative infinity, reaching a peak, descending through the origin, reaching a trough, then ascending to positive infinity. (b) is a line with negative slope. (c) is a circle of radius 3 centered at the origin.

Applying the Vertical Line Test: Which graphs represent functions?

If any vertical line intersects a graph more than once, the relation represented by the graph is not a function. Notice that any vertical line would pass through only one point of the two graphs shown in graphs [latex](a)[/latex] and [latex](b)[/latex]. From this we can conclude that these two graphs represent functions. The third graph, [latex](c)[/latex], does not represent a function because, at most [latex]x[/latex]-values, a vertical line would intersect the graph at more than one point. This is shown in the diagram below.

A circle is graphed with a dotted vertical line crossing through two points of the circle, demonstrating that the circle does not pass the vertical line test and is not a function.

Not a Function: The vertical line test demonstrates that a circle is not a function.

Sours: https://courses.lumenlearning.com/boundless-algebra/chapter/introduction-to-functions/
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Functions and linear equations

If we in the following equation y=x+7 assigns a value to x, the equation will give us a value for y.


Example

$$y=x+7$$

$$if\; x=2\; then$$

$$y=2+7=9$$

If we would have assigned a different value for x, the equation would have given us another value for y. We could instead have assigned a value for y and solved the equation to find the matching value of x.

In our equation y=x+7, we have two variables, x and y. The variable which we assign the value we call the independent variable, and the other variable is the dependent variable, since it value depends on the independent variable. In our example above, x is the independent variable and y is the dependent variable.

A function is an equation that has only one answer for y for every x. A function assigns exactly one output to each input of a specified type.

It is common to name a function either f(x) or g(x) instead of y. f(2) means that we should find the value of our function when x equals 2.


Example

$$f(x)=x+7$$

$$if\; x=2\; then$$

$$f(2)=2+7=9$$

A function is linear if it can be defined by

$$f(x)=mx+b$$

f(x) is the value of the function.
m is the slope of the line.
b is the value of the function when x equals zero or the y-coordinate of the point where the line crosses the y-axis in the coordinate plane.
x is the value of the x-coordinate.

This form is called the slope-intercept form. If m, the slope, is negative the functions value decreases with an increasing x and the opposite if we have a positive slope.

An equation such as y=x+7 is linear and there are an infinite number of ordered pairs of x and y that satisfy the equation.

Graph 1

The slope, m, is here 1 and our b (y-intercept) is 7.
The slope of a line passing through points (x1,y1) and (x2,y2) is given by

$$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$

$$x_{2}\neq x_{1}$$

If two linear equations are given the same slope it means that they are parallel and if the product of two slopes m1*m2=-1 the two linear equations are said to be perpendicular.


Video lesson

If x is -1 what is the value for f(x) when f(x)=3x+5?

Sours: https://www.mathplanet.com/education/algebra-2/how-to-graph-functions-and-linear-equations/functions-and-linear-equations

Algebraic function

In mathematics, an algebraic function is a function that can be defined as the root of a polynomial equation. Quite often algebraic functions are algebraic expressions using a finite number of terms, involving only the algebraic operations addition, subtraction, multiplication, division, and raising to a fractional power. Examples of such functions are:

  • f(x) = 1/x
  • f(x)={\sqrt {x}}
  • {\displaystyle f(x)={\frac {\sqrt {1+x^{3}}}{x^{3/7}-{\sqrt {7}}x^{1/3}}}}

Some algebraic functions, however, cannot be expressed by such finite expressions (this is the Abel–Ruffini theorem). This is the case, for example, for the Bring radical, which is the function implicitly defined by

{\displaystyle f(x)^{5}+f(x)+x=0}.

In more precise terms, an algebraic function of degree n in one variable x is a function {\displaystyle y=f(x),} that is continuous in its domain and satisfies a polynomial equation

a_{n}(x)y^{n}+a_{n-1}(x)y^{n-1}+\cdots +a_{0}(x)=0

where the coefficients ai(x) are polynomial functions of x, with integer coefficients. It can be shown that the same class of functions is obtained if algebraic numbers are accepted for the coefficients of the ai(x)'s. If transcendental numbers occur in the coefficients the function is, in general, not algebraic, but it is algebraic over the field generated by these coefficients.

The value of an algebraic function at a rational number, and more generally, at an algebraic number is always an algebraic number. Sometimes, coefficients a_{i}(x) that are polynomial over a ringR are considered, and one then talks about "functions algebraic over R".

A function which is not algebraic is called a transcendental function, as it is for example the case of {\displaystyle \exp x,\tan x,\ln x,\Gamma (x)}. A composition of transcendental functions can give an algebraic function: {\displaystyle f(x)=\cos \arcsin x={\sqrt {1-x^{2}}}}.

As a polynomial equation of degreen has up to n roots (and exactly n roots over an algebraically closed field, such as the complex numbers), a polynomial equation does not implicitly define a single function, but up to n functions, sometimes also called branches. Consider for example the equation of the unit circle: y^{2}+x^{2}=1.\, This determines y, except only up to an overall sign; accordingly, it has two branches: y=\pm {\sqrt {1-x^{2}}}.\,

An algebraic function in m variables is similarly defined as a function {\displaystyle y=f(x_{1},\dots ,x_{m})} which solves a polynomial equation in m + 1 variables:

{\displaystyle p(y,x_{1},x_{2},\dots ,x_{m})=0.}

It is normally assumed that p should be an irreducible polynomial. The existence of an algebraic function is then guaranteed by the implicit function theorem.

Formally, an algebraic function in m variables over the field K is an element of the algebraic closure of the field of rational functionsK(x1, ..., xm).

Algebraic functions in one variable[edit]

Introduction and overview[edit]

The informal definition of an algebraic function provides a number of clues about their properties. To gain an intuitive understanding, it may be helpful to regard algebraic functions as functions which can be formed by the usual algebraic operations: addition, multiplication, division, and taking an nth root. This is something of an oversimplification; because of the fundamental theorem of Galois theory, algebraic functions need not be expressible by radicals.

First, note that any polynomial functiony=p(x) is an algebraic function, since it is simply the solution y to the equation

y-p(x)=0.\,

More generally, any rational functiony={\frac {p(x)}{q(x)}} is algebraic, being the solution to

q(x)y-p(x)=0.

Moreover, the nth root of any polynomial {\textstyle y={\sqrt[{n}]{p(x)}}} is an algebraic function, solving the equation

y^{n}-p(x)=0.

Surprisingly, the inverse function of an algebraic function is an algebraic function. For supposing that y is a solution to

a_{n}(x)y^{n}+\cdots +a_{0}(x)=0,

for each value of x, then x is also a solution of this equation for each value of y. Indeed, interchanging the roles of x and y and gathering terms,

b_{m}(y)x^{m}+b_{m-1}(y)x^{m-1}+\cdots +b_{0}(y)=0.

Writing x as a function of y gives the inverse function, also an algebraic function.

However, not every function has an inverse. For example, y = x2 fails the horizontal line test: it fails to be one-to-one. The inverse is the algebraic "function" {\displaystyle x=\pm {\sqrt {y}}}. Another way to understand this, is that the set of branches of the polynomial equation defining our algebraic function is the graph of an algebraic curve.

The role of complex numbers[edit]

From an algebraic perspective, complex numbers enter quite naturally into the study of algebraic functions. First of all, by the fundamental theorem of algebra, the complex numbers are an algebraically closed field. Hence any polynomial relation p(y, x) = 0 is guaranteed to have at least one solution (and in general a number of solutions not exceeding the degree of p in y) for y at each point x, provided we allow y to assume complex as well as real values. Thus, problems to do with the domain of an algebraic function can safely be minimized.

A graph of three branches of the algebraic function y, where y3 − xy + 1 = 0, over the domain 3/22/3< x< 50.

Furthermore, even if one is ultimately interested in real algebraic functions, there may be no means to express the function in terms of addition, multiplication, division and taking nth roots without resorting to complex numbers (see casus irreducibilis). For example, consider the algebraic function determined by the equation

y^{3}-xy+1=0.\,

Using the cubic formula, we get

y=-{\frac {2x}{\sqrt[{3}]{-108+12{\sqrt {81-12x^{3}}}}}}+{\frac {\sqrt[{3}]{-108+12{\sqrt {81-12x^{3}}}}}{6}}.

For x\leq {\frac {3}{\sqrt[{3}]{4}}}, the square root is real and the cubic root is thus well defined, providing the unique real root. On the other hand, for x>{\frac {3}{\sqrt[{3}]{4}}}, the square root is not real, and one has to choose, for the square root, either non-real square root. Thus the cubic root has to be chosen among three non-real numbers. If the same choices are done in the two terms of the formula, the three choices for the cubic root provide the three branches shown, in the accompanying image.

It may be proven that there is no way to express this function in terms of nth roots using real numbers only, even though the resulting function is real-valued on the domain of the graph shown.

On a more significant theoretical level, using complex numbers allows one to use the powerful techniques of complex analysis to discuss algebraic functions. In particular, the argument principle can be used to show that any algebraic function is in fact an analytic function, at least in the multiple-valued sense.

Formally, let p(x, y) be a complex polynomial in the complex variables x and y. Suppose that x0 ∈ C is such that the polynomial p(x0, y) of y has n distinct zeros. We shall show that the algebraic function is analytic in a neighborhood of x0. Choose a system of n non-overlapping discs Δi containing each of these zeros. Then by the argument principle

{\frac {1}{2\pi i}}\oint _{\partial \Delta _{i}}{\frac {p_{y}(x_{0},y)}{p(x_{0},y)}}\,dy=1.

By continuity, this also holds for all x in a neighborhood of x0. In particular, p(x, y) has only one root in Δi, given by the residue theorem:

f_{i}(x)={\frac {1}{2\pi i}}\oint _{\partial \Delta _{i}}y{\frac {p_{y}(x,y)}{p(x,y)}}\,dy

which is an analytic function.

Monodromy[edit]

Note that the foregoing proof of analyticity derived an expression for a system of n different function elementsfi(x), provided that x is not a critical point of p(x, y). A critical point is a point where the number of distinct zeros is smaller than the degree of p, and this occurs only where the highest degree term of p vanishes, and where the discriminant vanishes. Hence there are only finitely many such points c1, ..., cm.

A close analysis of the properties of the function elements fi near the critical points can be used to show that the monodromy cover is ramified over the critical points (and possibly the point at infinity). Thus the holomorphic extension of the fi has at worst algebraic poles and ordinary algebraic branchings over the critical points.

Note that, away from the critical points, we have

p(x,y)=a_{n}(x)(y-f_{1}(x))(y-f_{2}(x))\cdots (y-f_{n}(x))

since the fi are by definition the distinct zeros of p. The monodromy group acts by permuting the factors, and thus forms the monodromy representation of the Galois group of p. (The monodromy action on the universal covering space is related but different notion in the theory of Riemann surfaces.)

History[edit]

The ideas surrounding algebraic functions go back at least as far as René Descartes. The first discussion of algebraic functions appears to have been in Edward Waring's 1794 An Essay on the Principles of Human Knowledge in which he writes:

let a quantity denoting the ordinate, be an algebraic function of the abscissa x, by the common methods of division and extraction of roots, reduce it into an infinite series ascending or descending according to the dimensions of x, and then find the integral of each of the resulting terms.

See also[edit]

References[edit]

External links[edit]

Sours: https://en.wikipedia.org/wiki/Algebraic_function

Algebra functions

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Section 3-4 : The Definition of a Function

We now need to move into the second topic of this chapter. Before we do that however we need a quick definition taken care of.

Definition of Relation

A relation is a set of ordered pairs.

This seems like an odd definition but we’ll need it for the definition of a function (which is the main topic of this section). However, before we actually give the definition of a function let’s see if we can get a handle on just what a relation is.

Think back to Example 1 in the Graphing section of this chapter. In that example we constructed a set of ordered pairs we used to sketch the graph of \(y = {\left( {x - 1} \right)^2} - 4\). Here are the ordered pairs that we used.

\[\left( { - 2,5} \right)\,\,\,\,\left( { - 1,0} \right)\,\,\,\,\left( {0, - 3} \right)\,\,\,\,\left( {1, - 4} \right)\,\,\,\,\left( {2, - 3} \right)\,\,\,\,\left( {3,0} \right)\,\,\,\,\left( {4,5} \right)\]

Any of the following are then relations because they consist of a set of ordered pairs.

\[\begin{align*}& \left\{ {\left( { - 2,5} \right)\,\,\,\,\left( { - 1,0} \right)\,\,\,\,\left( {2, - 3} \right)} \right\}\\ & \left\{ {\left( { - 1,0} \right)\,\,\,\,\left( {0, - 3} \right)\,\,\,\,\left( {2, - 3} \right)\,\,\,\,\left( {3,0} \right)\,\,\,\,\left( {4,5} \right)} \right\}\\ & \left\{ {\left( {3,0} \right)\,\,\,\,\left( {4,5} \right)} \right\}\\ & \left\{ {\left( { - 2,5} \right)\,\,\,\,\left( { - 1,0} \right)\,\,\,\,\left( {0, - 3} \right)\,\,\,\,\left( {1, - 4} \right)\,\,\,\,\left( {2, - 3} \right)\,\,\,\,\left( {3,0} \right)\,\,\,\,\left( {4,5} \right)} \right\}\end{align*}\]

There are of course many more relations that we could form from the list of ordered pairs above, but we just wanted to list a few possible relations to give some examples. Note as well that we could also get other ordered pairs from the equation and add those into any of the relations above if we wanted to.

Now, at this point you are probably asking just why we care about relations and that is a good question. Some relations are very special and are used at almost all levels of mathematics. The following definition tells us just which relations are these special relations.

Definition of a Function

A function is a relation for which each value from the set the first components of the ordered pairs is associated with exactly one value from the set of second components of the ordered pair.

Okay, that is a mouth full. Let’s see if we can figure out just what it means. Let’s take a look at the following example that will hopefully help us figure all this out.

Example 1The following relation is a function. \[\left\{ {\left( { - 1,0} \right)\,\,\,\,\left( {0, - 3} \right)\,\,\,\,\left( {2, - 3} \right)\,\,\,\,\left( {3,0} \right)\,\,\,\,\left( {4,5} \right)} \right\}\]
Show Solution

From these ordered pairs we have the following sets of first components (i.e. the first number from each ordered pair) and second components (i.e. the second number from each ordered pair).

\[{1^{{\mbox{st}}}}{\mbox{ components : }}\left\{ { - 1,0,2,3,4} \right\}\hspace{0.25in}\hspace{0.25in}{2^{{\mbox{nd}}}}{\mbox{ components : }}\left\{ {0, - 3,0,5} \right\}\]

For the set of second components notice that the “-3” occurred in two ordered pairs but we only listed it once.

To see why this relation is a function simply pick any value from the set of first components. Now, go back up to the relation and find every ordered pair in which this number is the first component and list all the second components from those ordered pairs. The list of second components will consist of exactly one value.

For example, let’s choose 2 from the set of first components. From the relation we see that there is exactly one ordered pair with 2 as a first component,\(\left( {2, - 3} \right)\). Therefore, the list of second components (i.e. the list of values from the set of second components) associated with 2 is exactly one number, -3.

Note that we don’t care that -3 is the second component of a second ordered par in the relation. That is perfectly acceptable. We just don’t want there to be any more than one ordered pair with 2 as a first component.

We looked at a single value from the set of first components for our quick example here but the result will be the same for all the other choices. Regardless of the choice of first components there will be exactly one second component associated with it.

Therefore, this relation is a function.

In order to really get a feel for what the definition of a function is telling us we should probably also check out an example of a relation that is not a function.

Example 2The following relation is not a function. \[\left\{ {\left( {6,10} \right)\,\,\,\,\left( { - 7,3} \right)\,\,\,\,\left( {0,4} \right)\,\,\,\,\left( {6, - 4} \right)} \right\}\]
Show Solution

Don’t worry about where this relation came from. It is just one that we made up for this example.

Here is the list of first and second components

\[{1^{{\mbox{st}}}}{\mbox{ components : }}\left\{ {6, - 7,0} \right\}\hspace{0.25in}\hspace{0.25in}{2^{{\mbox{nd}}}}{\mbox{ components : }}\left\{ {10,3,4, - 4} \right\}\]

From the set of first components let’s choose 6. Now, if we go up to the relation we see that there are two ordered pairs with 6 as a first component : \(\left( {6,10} \right)\) and \(\left( {6, - 4} \right)\). The list of second components associated with 6 is then : 10, -4.

The list of second components associated with 6 has two values and so this relation is not a function.

Note that the fact that if we’d chosen -7 or 0 from the set of first components there is only one number in the list of second components associated with each. This doesn’t matter. The fact that we found even a single value in the set of first components with more than one second component associated with it is enough to say that this relation is not a function.

As a final comment about this example let’s note that if we removed the first and/or the fourth ordered pair from the relation we would have a function!

So, hopefully you have at least a feeling for what the definition of a function is telling us.

Now that we’ve forced you to go through the actual definition of a function let’s give another “working” definition of a function that will be much more useful to what we are doing here.

The actual definition works on a relation. However, as we saw with the four relations we gave prior to the definition of a function and the relation we used in Example 1 we often get the relations from some equation.

It is important to note that not all relations come from equations! The relation from the second example for instance was just a set of ordered pairs we wrote down for the example and didn’t come from any equation. This can also be true with relations that are functions. They do not have to come from equations.

However, having said that, the functions that we are going to be using in this course do all come from equations. Therefore, let’s write down a definition of a function that acknowledges this fact.

Before we give the “working” definition of a function we need to point out that this is NOT the actual definition of a function, that is given above. This is simply a good “working definition” of a function that ties things to the kinds of functions that we will be working with in this course.

“Working Definition” of Function

A function is an equation for which any \(x\) that can be plugged into the equation will yield exactly one \(y\) out of the equation.

There it is. That is the definition of functions that we’re going to use and will probably be easier to decipher just what it means.

Before we examine this a little more note that we used the phrase “\(x\) that can be plugged into” in the definition. This tends to imply that not all \(x\)’s can be plugged into an equation and this is in fact correct. We will come back and discuss this in more detail towards the end of this section, however at this point just remember that we can’t divide by zero and if we want real numbers out of the equation we can’t take the square root of a negative number. So, with these two examples it is clear that we will not always be able to plug in every \(x\) into any equation.

Further, when dealing with functions we are always going to assume that both \(x\) and \(y\) will be real numbers. In other words, we are going to forget that we know anything about complex numbers for a little bit while we deal with this section.

Okay, with that out of the way let’s get back to the definition of a function and let’s look at some examples of equations that are functions and equations that aren’t functions.

Example 3Determine which of the following equations are functions and which are not functions.
  1. \(y = 5x + 1\)
  2. \(y = {x^2} + 1\)
  3. \({y^2} = x + 1\)
  4. \({x^2} + {y^2} = 4\)
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Show Discussion

The “working” definition of function is saying is that if we take all possible values of \(x\) and plug them into the equation and solve for \(y\) we will get exactly one value for each value of \(x\). At this stage of the game it can be pretty difficult to actually show that an equation is a function so we’ll mostly talk our way through it. On the other hand, it’s often quite easy to show that an equation isn’t a function.


a\(y = 5x + 1\) Show Solution

So, we need to show that no matter what \(x\) we plug into the equation and solve for \(y\) we will only get a single value of \(y\). Note as well that the value of \(y\) will probably be different for each value of \(x\), although it doesn’t have to be.

Let’s start this off by plugging in some values of \(x\) and see what happens.

\[\begin{align*}x & = - 4:\hspace{0.25in} & y & = 5\left( { - 4} \right) + 1 = - 20 + 1 = - 19\\ x & = 0:\hspace{0.25in} & y & = 5\left( 0 \right) + 1 = 0 + 1 = 1\\ x & = 10:\hspace{0.25in} & y & = 5\left( {10} \right) + 1 = 50 + 1 = 51\end{align*}\]

So, for each of these values of \(x\) we got a single value of \(y\) out of the equation. Now, this isn’t sufficient to claim that this is a function. In order to officially prove that this is a function we need to show that this will work no matter which value of \(x\) we plug into the equation.

Of course, we can’t plug all possible value of \(x\) into the equation. That just isn’t physically possible. However, let’s go back and look at the ones that we did plug in. For each \(x\), upon plugging in, we first multiplied the \(x\) by 5 and then added 1 onto it. Now, if we multiply a number by 5 we will get a single value from the multiplication. Likewise, we will only get a single value if we add 1 onto a number. Therefore, it seems plausible that based on the operations involved with plugging \(x\) into the equation that we will only get a single value of \(y\) out of the equation.

So, this equation is a function.


b\(y = {x^2} + 1\) Show Solution

Again, let’s plug in a couple of values of \(x\) and solve for \(y\) to see what happens.

\[\begin{align*}x & = - 1:\hspace{0.25in} & y = {\left( { - 1} \right)^2} + 1 = 1 + 1 = 2\\ x & = 3:\hspace{0.25in} & y = {\left( 3 \right)^2} + 1 = 9 + 1 = 10\end{align*}\]

Now, let’s think a little bit about what we were doing with the evaluations. First, we squared the value of \(x\) that we plugged in. When we square a number there will only be one possible value. We then add 1 onto this, but again, this will yield a single value.

So, it seems like this equation is also a function.

Note that it is okay to get the same \(y\) value for different \(x\)’s. For example,

\[x = - 3:\hspace{0.25in}y = {\left( { - 3} \right)^2} + 1 = 9 + 1 = 10\]

We just can’t get more than one \(y\) out of the equation after we plug in the \(x\).


c\({y^2} = x + 1\) Show Solution

As we’ve done with the previous two equations let’s plug in a couple of value of \(x\), solve for \(y\) and see what we get.

\[\begin{align*}x & = 3:\hspace{0.25in} & {y^2} & = 3 + 1 = 4\hspace{0.25in}\Rightarrow \hspace{0.25in} y = \pm 2\\ x & = - 1:\hspace{0.25in} & {y^2} & = - 1 + 1 = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}y = 0\\ x & = 10:\hspace{0.25in} & {y^2} & = 10 + 1 = 11\hspace{0.25in} \Rightarrow \hspace{0.25in}y = \pm \sqrt {11} \end{align*}\]

Now, remember that we’re solving for \(y\) and so that means that in the first and last case above we will actually get two different \(y\) values out of the \(x\) and so this equation is NOT a function.

Note that we can have values of \(x\) that will yield a single \(y\) as we’ve seen above, but that doesn’t matter. If even one value of \(x\) yields more than one value of \(y\) upon solving the equation will not be a function.

What this really means is that we didn’t need to go any farther than the first evaluation, since that gave multiple values of \(y\).


d\({x^2} + {y^2} = 4\) Show Solution

With this case we’ll use the lesson learned in the previous part and see if we can find a value of \(x\) that will give more than one value of \(y\) upon solving. Because we’ve got a y2 in the problem this shouldn’t be too hard to do since solving will eventually mean using the square root property which will give more than one value of \(y\).

\[x = 0:\hspace{0.25in}{0^2} + {y^2} = 4\hspace{0.25in} \Rightarrow \hspace{0.25in}{y^2} = 4\hspace{0.25in} \Rightarrow \hspace{0.25in} y = \pm \,2\]

So, this equation is not a function. Recall, that from the previous section this is the equation of a circle. Circles are never functions.

Hopefully these examples have given you a better feel for what a function actually is.

We now need to move onto something called function notation. Function notation will be used heavily throughout most of the remaining chapters in this course and so it is important to understand it.

Let’s start off with the following quadratic equation.

\[y = {x^2} - 5x + 3\]

We can use a process similar to what we used in the previous set of examples to convince ourselves that this is a function. Since this is a function we will denote it as follows,

\[f\left( x \right) = {x^2} - 5x + 3\]

So, we replaced the \(y\) with the notation \(f\left( x \right)\). This is read as “f of \(x\)”. Note that there is nothing special about the \(f\) we used here. We could just have easily used any of the following,

\[g\left( x \right) = {x^2} - 5x + 3\,\,\,\hspace{0.25in}h\left( x \right) = {x^2} - 5x + 3\hspace{0.25in}R\left( x \right) = {x^2} - 5x + 3\]

The letter we use does not matter. What is important is the “\(\left( x \right)\)” part. The letter in the parenthesis must match the variable used on the right side of the equal sign.

It is very important to note that \(f\left( x \right)\) is really nothing more than a really fancy way of writing \(y\). If you keep that in mind you may find that dealing with function notation becomes a little easier.

Also, this is NOT a multiplication of \(f\) by \(x\)! This is one of the more common mistakes people make when they first deal with functions. This is just a notation used to denote functions.

Next we need to talk about evaluating functions. Evaluating a function is really nothing more than asking what its value is for specific values of \(x\). Another way of looking at it is that we are asking what the \(y\) value for a given \(x\) is.

Evaluation is really quite simple. Let’s take the function we were looking at above

\[f\left( x \right) = {x^2} - 5x + 3\]

and ask what its value is for \(x = 4\). In terms of function notation we will “ask” this using the notation \(f\left( 4 \right)\). So, when there is something other than the variable inside the parenthesis we are really asking what the value of the function is for that particular quantity.

Now, when we say the value of the function we are really asking what the value of the equation is for that particular value of \(x\). Here is \(f\left( 4 \right)\).

\[f\left( 4 \right) = {\left( 4 \right)^2} - 5\left( 4 \right) + 3 = 16 - 20 + 3 = - 1\]

Notice that evaluating a function is done in exactly the same way in which we evaluate equations. All we do is plug in for \(x\) whatever is on the inside of the parenthesis on the left. Here’s another evaluation for this function.

\[f\left( { - 6} \right) = {\left( { - 6} \right)^2} - 5\left( { - 6} \right) + 3 = 36 + 30 + 3 = 69\]

So, again, whatever is on the inside of the parenthesis on the left is plugged in for \(x\) in the equation on the right. Let’s take a look at some more examples.

Example 4Given \(f\left( x \right) = {x^2} - 2x + 8\) and \(g\left( x \right) = \sqrt {x + 6} \) evaluate each of the following.
  1. \(f\left( 3 \right)\) and \(g\left( 3 \right)\)
  2. \(f\left( { - 10} \right)\) and \(g\left( { - 10} \right)\)
  3. \(f\left( 0 \right)\)
  4. \(f\left( t \right)\)
  5. \(f\left( {t + 1} \right)\) and \(f\left( {x + 1} \right)\)
  6. \(f\left( {{x^3}} \right)\)
  7. \(g\left( {{x^2} - 5} \right)\)
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a\(f\left( 3 \right)\) and \(g\left( 3 \right)\) Show Solution

Okay we’ve got two function evaluations to do here and we’ve also got two functions so we’re going to need to decide which function to use for the evaluations. The key here is to notice the letter that is in front of the parenthesis. For \(f\left( 3 \right)\) we will use the function \(f\left( x \right)\) and for \(g\left( 3 \right)\) we will use \(g\left( x \right)\). In other words, we just need to make sure that the variables match up.

Here are the evaluations for this part.

\[\begin{align*}f\left( 3 \right) & = {\left( 3 \right)^2} - 2\left( 3 \right) + 8 = 9 - 6 + 8 = 11\\ g\left( 3 \right) & = \sqrt {3 + 6} = \sqrt 9 = 3\end{align*}\]

b\(f\left( { - 10} \right)\) and \(g\left( { - 10} \right)\) Show Solution

This one is pretty much the same as the previous part with one exception that we’ll touch on when we reach that point. Here are the evaluations.

\[f\left( { - 10} \right) = {\left( { - 10} \right)^2} - 2\left( { - 10} \right) + 8 = 100 + 20 + 8 = 128\]

Make sure that you deal with the negative signs properly here. Now the second one.

\[g\left( { - 10} \right) = \sqrt { - 10 + 6} = \sqrt { - 4} \]

We’ve now reached the difference. Recall that when we first started talking about the definition of functions we stated that we were only going to deal with real numbers. In other words, we only plug in real numbers and we only want real numbers back out as answers. So, since we would get a complex number out of this we can’t plug -10 into this function.


c\(f\left( 0 \right)\) Show Solution

Not much to this one.

\[f\left( 0 \right) = {\left( 0 \right)^2} - 2\left( 0 \right) + 8 = 8\]

Again, don’t forget that this isn’t multiplication! For some reason students like to think of this one as multiplication and get an answer of zero. Be careful.


d\(f\left( t \right)\) Show Solution

The rest of these evaluations are now going to be a little different. As this one shows we don’t need to just have numbers in the parenthesis. However, evaluation works in exactly the same way. We plug into the \(x\)’s on the right side of the equal sign whatever is in the parenthesis. In this case that means that we plug in \(t\) for all the \(x\)’s.

Here is this evaluation.

\[f\left( t \right) = {t^2} - 2t + 8\]

Note that in this case this is pretty much the same thing as our original function, except this time we’re using \(t\) as a variable.


e\(f\left( {t + 1} \right)\) and \(f\left( {x + 1} \right)\) Show Solution

Now, let’s get a little more complicated, or at least they appear to be more complicated. Things aren’t as bad as they may appear however. We’ll evaluate \(f\left( {t + 1} \right)\) first. This one works exactly the same as the previous part did. All the \(x\)’s on the left will get replaced with \(t + 1\). We will have some simplification to do as well after the substitution.

\[\begin{align*}f\left( {t + 1} \right) & = {\left( {t + 1} \right)^2} - 2\left( {t + 1} \right) + 8\\ & = {t^2} + 2t + 1 - 2t - 2 + 8\\ & = {t^2} + 7\end{align*}\]

Be careful with parenthesis in these kinds of evaluations. It is easy to mess up with them.

Now, let’s take a look at \(f\left( {x + 1} \right)\). With the exception of the \(x\) this is identical to \(f\left( {t + 1} \right)\) and so it works exactly the same way.

\[\begin{align*}f\left( {x + 1} \right) & = {\left( {x + 1} \right)^2} - 2\left( {x + 1} \right) + 8\\ & = {x^2} + 2x + 1 - 2x - 2 + 8\\ & = {x^2} + 7\end{align*}\]

Do not get excited about the fact that we reused \(x\)’s in the evaluation here. In many places where we will be doing this in later sections there will be \(x\)’s here and so you will need to get used to seeing that.


f\(f\left( {{x^3}} \right)\) Show Solution

Again, don’t get excited about the \(x\)’s in the parenthesis here. Just evaluate it as if it were a number.

\[\begin{align*}f\left( {{x^3}} \right) & = {\left( {{x^3}} \right)^2} - 2\left( {{x^3}} \right) + 8\\ & = {x^6} - 2{x^3} + 8\end{align*}\]

g\(g\left( {{x^2} - 5} \right)\) Show Solution

One more evaluation and this time we’ll use the other function.

\[\begin{align*}g\left( {{x^2} - 5} \right) & = \sqrt {{x^2} - 5 + 6} \\ & = \sqrt {{x^2} + 1} \end{align*}\]

Function evaluation is something that we’ll be doing a lot of in later sections and chapters so make sure that you can do it. You will find several later sections very difficult to understand and/or do the work in if you do not have a good grasp on how function evaluation works.

While we are on the subject of function evaluation we should now talk about piecewise functions. We’ve actually already seen an example of a piecewise function even if we didn’t call it a function (or a piecewise function) at the time. Recall the mathematical definition of absolute value.

\[\left| x \right| = \left\{ {\begin{array}{*{20}{l}}x&{{\mbox{if }}x \ge 0}\\{ - x}&{{\mbox{if }}x < 0}\end{array}} \right.\]

This is a function and if we use function notation we can write it as follows,

\[f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}x&{{\mbox{if }}x \ge 0}\\{ - x}&{{\mbox{if }}x < 0}\end{array}} \right.\]

This is also an example of a piecewise function. A piecewise function is nothing more than a function that is broken into pieces and which piece you use depends upon value of \(x\). So, in the absolute value example we will use the top piece if \(x\) is positive or zero and we will use the bottom piece if \(x\) is negative.

Let’s take a look at evaluating a more complicated piecewise function.

Example 5Given, \[g\left( t \right) = \left\{ {\begin{array}{*{20}{l}}{3{t^2} + 4}&{{\mbox{if }}t \le - 4}\\{10}&{{\mbox{if }} - 4 < t \le 15}\\{1 - 6t}&{{\mbox{if }}t > 15}\end{array}} \right.\]

evaluate each of the following.

  1. \(g\left( { - 6} \right)\)
  2. \(g\left( { - 4} \right)\)
  3. \(g\left( 1 \right)\)
  4. \(g\left( {15} \right)\)
  5. \(g\left( {21} \right)\)
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Show Discussion

Before starting the evaluations here let’s notice that we’re using different letters for the function and variable than the ones that we’ve used to this point. That won’t change how the evaluation works. Do not get so locked into seeing \(f\) for the function and \(x\) for the variable that you can’t do any problem that doesn’t have those letters.

Now, to do each of these evaluations the first thing that we need to do is determine which inequality the number satisfies, and it will only satisfy a single inequality. When we determine which inequality the number satisfies we use the equation associated with that inequality.

So, let’s do some evaluations.


a\(g\left( { - 6} \right)\) Show Solution

In this case -6 satisfies the top inequality and so we’ll use the top equation for this evaluation.

\[g\left( { - 6} \right) = 3{\left( { - 6} \right)^2} + 4 = 112\]

b\(g\left( { - 4} \right)\) Show Solution

Now we’ll need to be a little careful with this one since -4 shows up in two of the inequalities. However, it only satisfies the top inequality and so we will once again use the top function for the evaluation.

\[g\left( { - 4} \right) = 3{\left( { - 4} \right)^2} + 4 = 52\]

c\(g\left( 1 \right)\) Show Solution

In this case the number, 1, satisfies the middle inequality and so we’ll use the middle equation for the evaluation. This evaluation often causes problems for students despite the fact that it’s actually one of the easiest evaluations we’ll ever do. We know that we evaluate functions/equations by plugging in the number for the variable. In this case there are no variables. That isn’t a problem. Since there aren’t any variables it just means that we don’t actually plug in anything and we get the following,

\[g\left( 1 \right) = 10\]

d\(g\left( {15} \right)\) Show Solution

Again, like with the second part we need to be a little careful with this one. In this case the number satisfies the middle inequality since that is the one with the equal sign in it. Then like the previous part we just get,

\[g\left( {15} \right) = 10\]

Don’t get excited about the fact that the previous two evaluations were the same value. This will happen on occasion.


e\(g\left( {21} \right)\) Show Solution

For the final evaluation in this example the number satisfies the bottom inequality and so we’ll use the bottom equation for the evaluation.

\[g\left( {21} \right) = 1 - 6\left( {21} \right) = - 125\]

Piecewise functions do not arise all that often in an Algebra class however, they do arise in several places in later classes and so it is important for you to understand them if you are going to be moving on to more math classes.

As a final topic we need to come back and touch on the fact that we can’t always plug every \(x\) into every function. We talked briefly about this when we gave the definition of the function and we saw an example of this when we were evaluating functions. We now need to look at this in a little more detail.

First, we need to get a couple of definitions out of the way.

Domain and Range

The domain of an equation is the set of all \(x\)’s that we can plug into the equation and get back a real number for \(y\). The range of an equation is the set of all \(y\)’s that we can ever get out of the equation.

Note that we did mean to use equation in the definitions above instead of functions. These are really definitions for equations. However, since functions are also equations we can use the definitions for functions as well.

Determining the range of an equation/function can be pretty difficult to do for many functions and so we aren’t going to really get into that. We are much more interested here in determining the domains of functions. From the definition the domain is the set of all \(x\)’s that we can plug into a function and get back a real number. At this point, that means that we need to avoid division by zero and taking square roots of negative numbers.

Let’s do a couple of quick examples of finding domains.

Example 6Determine the domain of each of the following functions.
  1. \(\displaystyle g\left( x \right) = \frac{{x + 3}}{{{x^2} + 3x - 10}}\)
  2. \(f\left( x \right) = \sqrt {5 - 3x} \)
  3. \(\displaystyle h\left( x \right) = \frac{{\sqrt {7x + 8} }}{{{x^2} + 4}}\)
  4. \(\displaystyle R\left( x \right) = \frac{{\sqrt {10x - 5} }}{{{x^2} - 16}}\)
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The domains for these functions are all the values of \(x\) for which we don’t have division by zero or the square root of a negative number. If we remember these two ideas finding the domains will be pretty easy.


a\(\displaystyle g\left( x \right) = \frac{{x + 3}}{{{x^2} + 3x - 10}}\) Show Solution

So, in this case there are no square roots so we don’t need to worry about the square root of a negative number. There is however a possibility that we’ll have a division by zero error. To determine if we will we’ll need to set the denominator equal to zero and solve.

\[{x^2} + 3x - 10 = \left( {x + 5} \right)\left( {x - 2} \right) = 0\hspace{0.25in}x = - 5,\,\,x = 2\]

So, we will get division by zero if we plug in \(x = - 5\) or \(x = 2\). That means that we’ll need to avoid those two numbers. However, all the other values of \(x\) will work since they don’t give division by zero. The domain is then,

\[{\mbox{Domain : All real numbers except }}x = - 5{\mbox{ and }}x = 2\]

b\(f\left( x \right) = \sqrt {5 - 3x} \) Show Solution

In this case we won’t have division by zero problems since we don’t have any fractions. We do have a square root in the problem and so we’ll need to worry about taking the square root of a negative numbers.

This one is going to work a little differently from the previous part. In that part we determined the value(s) of \(x\) to avoid. In this case it will be just as easy to directly get the domain. To avoid square roots of negative numbers all that we need to do is require that

\[5 - 3x \ge 0\]

This is a fairly simple linear inequality that we should be able to solve at this point.

\[5 \ge 3x\hspace{0.25in} \Rightarrow \hspace{0.25in}x \le \frac{5}{3}\]

The domain of this function is then,

\[{\mbox{Domain : }}x \le \frac{5}{3}\]

c\(\displaystyle h\left( x \right) = \frac{{\sqrt {7x + 8} }}{{{x^2} + 4}}\) Show Solution

In this case we’ve got a fraction, but notice that the denominator will never be zero for any real number since x2 is guaranteed to be positive or zero and adding 4 onto this will mean that the denominator is always at least 4. In other words, the denominator won’t ever be zero. So, all we need to do then is worry about the square root in the numerator.

To do this we’ll require,

\[\begin{align*}7x + 8 & \ge 0\\ 7x & \ge - 8\\ x & \ge - \frac{8}{7}\end{align*}\]

Now, we can actually plug in any value of \(x\) into the denominator, however, since we’ve got the square root in the numerator we’ll have to make sure that all \(x\)’s satisfy the inequality above to avoid problems. Therefore, the domain of this function is

\[{\mbox{Domain :}}\,\,x \ge - \frac{8}{7}\]

d\(\displaystyle R\left( x \right) = \frac{{\sqrt {10x - 5} }}{{{x^2} - 16}}\) Show Solution

In this final part we’ve got both a square root and division by zero to worry about. Let’s take care of the square root first since this will probably put the largest restriction on the values of \(x\). So, to keep the square root happy (i.e. no square root of negative numbers) we’ll need to require that,

\[\begin{align*}10x - 5 & \ge 0\\ 10x & \ge 5\\ x & \ge \frac{1}{2}\end{align*}\]

So, at the least we’ll need to require that \(x \ge \frac{1}{2}\) in order to avoid problems with the square root.

Now, let’s see if we have any division by zero problems. Again, to do this simply set the denominator equal to zero and solve.

\[{x^2} - 16 = \left( {x - 4} \right)\left( {x + 4} \right) = 0\hspace{0.25in} \Rightarrow \hspace{0.25in} x = - 4,\,\,x = 4\]

Now, notice that \(x = - 4\) doesn’t satisfy the inequality we need for the square root and so that value of \(x\) has already been excluded by the square root. On the other hand, \(x = 4\) does satisfy the inequality. This means that it is okay to plug \(x = 4\) into the square root, however, since it would give division by zero we will need to avoid it.

The domain for this function is then,

\[{\mbox{Domain : }}x \ge \frac{1}{2}{\mbox{ except }}x = 4\]
Sours: https://tutorial.math.lamar.edu/classes/alg/functiondefn.aspx

You are in bliss. I move the zipper of your dressing gown. lower your dressing gown. you dont have anything else. naked.

Now discussing:

Said Klavdia Pavlovna a little thoughtfully. It was crowded on the platforms covered with a glass dome in Vienna's central station. The train that arrived fifteen minutes ago from Budapest was preparing to continue its journey to Munich, and then to Strasbourg.



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