# 3.2: Function Notation

Difficulty Level: At Grade Created by: CK-12

## What is Function Notation?

Objectives

The lesson objective for Function Notation is:

• Understanding function notation.

Introduction

If you think of the relationship between two quantities, you can think of this relationship in terms of a ‘function machine’. The function machine shows how the function responds to various functions. If I triple the input and subtract one, the machine will respond to the function in such a way that it will convert \begin{align*}x\end{align*} into \begin{align*}3x - 1\end{align*}. If the function is \begin{align*}f\end{align*}, and 3 is fed into the machine, \begin{align*}3(3) - 1 = 8\end{align*} comes out.

Watch This

Guidance

To represent functions as equations, the symbol \begin{align*}f(x)\end{align*} is often used. The symbol \begin{align*}f(x)\end{align*} is pronounced as “\begin{align*}f\end{align*} of \begin{align*}x\end{align*}.” This means that the equation is a function that is written in the terms of the variable \begin{align*}x\end{align*}. An example of such a function is \begin{align*}f(x) = 3x+4\end{align*}. Functions can also be written using a letter other than \begin{align*}f\end{align*} and a variable other than \begin{align*}x\end{align*}. An example of such a case is \begin{align*}v(t) = 2t^2 - 5\end{align*} and \begin{align*}d(h) = 4h-3\end{align*}. In addition to representing functions as equations, you have been shown a variety of ways by which a function can be represented. These ways included:

• As a graph
• As ordered pairs
• As a table of values
• As an arrow or mapping diagram
• As mapping notation

When a function is represented as an equation, an ordered pair can be determined by evaluating various values of the assigned variable. If \begin{align*}f(x)=3x-4\end{align*}, then \begin{align*}f(4)=?\end{align*}

\begin{align*}f(x) &= 3x-4\\ f(x) & = 3(4) - 4\\ f(4) & = 12-4\\ f(4) & = 8\end{align*}

Graphically, if \begin{align*}f(4) = 8\end{align*}, this means that the point (4, 8) is a point on the graph of the line.

Example A

If \begin{align*}f(x) = x^2 + 2x +5\end{align*} find.

a) \begin{align*}f(2)\end{align*}

b) \begin{align*}f(-7)\end{align*}

c) \begin{align*}f(1.4)\end{align*}

To determine the value of the function for the assigned values of the variable, substitute the values into the function.

\begin{align*}& f(x) = x^2 + 2x+5 && \quad f(x) = x^2+2x+5 && \quad f(x)=x^2+2x+5\\ & {\color{red}\downarrow} \qquad \ \ {\color{red}\downarrow} \qquad {\color{red}\searrow} && \quad \ {\color{red}\downarrow} \qquad \ \ {\color{red}\downarrow} \qquad \ {\color{red}\searrow} && \quad \ \ {\color{red}\downarrow} \qquad \ {\color{red}\downarrow} \qquad \ {\color{red}\searrow}\\ & f(2) =(2)^2 +2(2) + 5 && \ f(-7) = (-7)^2+2(-7)+5 && \ f(1.4) = (1.4)^2+2(1.4) + 5\\ & f(2) = 4 + 4 + 5 && \ f(-7) = 49 - 14 +5 && \ f(1.4)=1.96 +2.8+5\\ & \boxed{f(2)=13} && \boxed{f(-7)=40} && \boxed{f(1.4) = 9.76}\end{align*}

Example B

Functions can also be represented as mapping rules. If \begin{align*}g(x)\rightarrow 5-2x\end{align*} find the following in simplest form:

a) \begin{align*}g(y)\end{align*}

b) \begin{align*}g(y-3)\end{align*}

c) \begin{align*}g(2y)\end{align*}

\begin{align*}& \ g(x) \rightarrow 5 -2x && \qquad g(x) \rightarrow 5-2x \mapsto\\ & \ {\color{red}\downarrow} \qquad \qquad \quad {\color{red}\downarrow} && \qquad \ \ {\color{red}\downarrow} \qquad \qquad \ {\color{red}\downarrow}\\ & \ g(y) \rightarrow 5 - 2(y) && \ g(y-3) \rightarrow 5-2 (y-3) && \text{Apply the distributive property}\\ & \boxed{g(y) \rightarrow 5 - 2y} && \ g(y-3) \rightarrow 5-2y+6 && \text{Combine like terms}\\ & && \boxed{g(y-3) \rightarrow 11-2y}\\ \\ & \ \ g(x) \rightarrow 5-2x\\ & \ g(2y) \rightarrow 5-2(2y)\\ & \boxed{g(2y) \rightarrow 5-4y}\end{align*}

Example C

If \begin{align*}P(a)=\frac{2a-3}{a+2}\end{align*}

a) evaluate

i) \begin{align*}P(0)\end{align*}

ii) \begin{align*}P(1)\end{align*}

iii) \begin{align*}P \left ( -\frac{1}{2} \right )\end{align*}

b) Find a value of ‘\begin{align*}a\end{align*}’ where \begin{align*}P(a)\end{align*} does not exist.

c) Find \begin{align*}P(a-2)\end{align*} in simplest form

d) Find ‘\begin{align*}a\end{align*}’ if \begin{align*}P(a)=-5\end{align*}

a) \begin{align*}& \ P(a) = \frac{2a-3}{a+2} && \ P(a) =\frac{2a-3}{a+2} && \qquad \ P(a)=\frac{2a-3}{a+2}\\ & \ P(0) =\frac{2(0)-3}{(0)+2} && \ P(1) = \frac{2(1)-3}{(1)+2} && \ P\left ( -\frac{1}{2} \right ) = \frac{2\left( -\frac{1}{2} \right )-3}{\left ( -\frac{1}{2} \right ) + 2}\\ & \boxed{P(0) = \frac{-3}{+2}} && \ P(1) = \frac{2-3}{1+2} && \ P \left ( -\frac{1}{2} \right ) = \frac{^1\cancel{2}\left ( -\frac{1}{\cancel{2}} \right )-3}{-\frac{1}{2} + \frac{4}{2}}\\ & && \boxed{P(1)=\frac{-1}{3}} && \ \ P \left ( -\frac{1}{2} \right ) = \frac{-1-3}{\frac{3}{2}}\\ & && && \ P\left ( -\frac{1}{2} \right ) = -4 \div \frac{3}{2}\\ & && && \ P \left ( -\frac{1}{2} \right ) = -4\left ( \frac{2}{3} \right )\\ & && && \boxed{P\left ( -\frac{1}{2} \right )} = \frac{-8}{3}\end{align*}

b) \begin{align*}P(a) = \frac{2a-3}{a+2}\end{align*} The function will not exist if the denominator equals zero.

\begin{align*}& \quad \ \ a+ 2 = 0\\ & a+2-2=0-2\\ & \qquad \quad \ \boxed{a=-2}\end{align*}

\begin{align*}& P(a) = \frac{2a-3}{(-2)+2}\\ & P(a) = \frac{2a-3}{0} && \text{Division by zero is undefined.}\end{align*}

Therefore, if \begin{align*}a=-2\end{align*}, then \begin{align*}P(a)=\frac{2a-3}{a+2}\end{align*} does not exist.

c) \begin{align*}& \qquad P(a) = \frac{2a-3}{a+2}\\ & \ P(a-2) = \frac{2(a-2)-3}{(a-2)+2} && \text{Substitue } a-2' \text{ for } a'\\ & \ P(a-2) = \frac{2a-4-3}{a-2+2} && \text{Remove brackets}\\ & \ P(a-2) = \frac{2a-7}{a} && \text{Combine like terms}\\ & \ P(a-2) = \frac{2\cancel{a}}{\cancel{a}} - \frac{7}{a} && \text{Express the fraction as two separate fractions and reduce.}\\ & \boxed{P(a-2) = 2-\frac{7}{a}}\end{align*}

d) \begin{align*}& \qquad \qquad \quad P(a) = \frac{2a-3}{a+2}\\ & \qquad \qquad \quad \ -5 = \frac{2a-3}{a+2} && \text{Let } P(a) = -5\\ & \qquad \ -5(a+2) = \left ( \frac{2a-3}{a+2} \right )(a+2) && \text{Multiply both sides by } (a+2)\\ & \qquad \ -5a -10 = \left ( \frac{2a-3}{\cancel{a+2}} \right ) (\cancel{a+2}) && \text{Simplify}\\ & \qquad \ -5a -10 = 2a-3 && \text{Solve the linear equation}\\ & -5a -10 -2a = 2a-2a-3 && \text{Move } 2a \text{ to the left by subtracting}\\ & \qquad \ -7a-10 = -3 && \text{Simplify}\\ & -7a-10+10 = -3+10 && \text{Move 10 to the right side by addition}\\ & \qquad \qquad \ -7a = 7 && \text{Simplify}\\ & \qquad \qquad \ \ \frac{-7a}{-7} = \frac{7}{-7} && \text{Divide both sides by -7 to solve for } a'.\\ & \qquad \qquad \qquad \boxed{a=-1}\end{align*}

Example D

Functions can also be used to represent and to solve word problems. The following is an example of this application.

The value \begin{align*}V\end{align*} of a digital camera \begin{align*}t\end{align*} years after it was bought is represented by the function \begin{align*}V(t) = 875 - 50t\end{align*}

a) Determine the value of \begin{align*}V(4)\end{align*} and explain what the solution mean to this problem.

b) Determine the value of \begin{align*}t\end{align*} then \begin{align*}V(t) = 525\end{align*} and explain what this represents.

c) What was the original cost of the digital camera?

a) \begin{align*}& \ V(t) = 875 - 50t\\ & \ V(4) = 875 - 50(4)\\ & \ V(4) = 875-200\\ & \boxed{V(4) = \ 675}\end{align*}

This means that the camera is valued at 675, 4 years after it was purchased. b) \begin{align*}& \qquad \ V(t) = 875 - 50t && \text{Let } V(t) = 525\\ & \qquad \ \ 525 = 875-50t && \text{Solve the equation}\\ & 525 -875 = 875 - 875 - 50t\\ & \quad \ -350 = - 50t\\ & \quad \ \ \frac{-350}{-50} = \frac{-50t}{-50}\\ & \qquad \quad \ \boxed{7 = t}\end{align*} The digital camera has a value of525, 7 years after it was purchased.

c) \begin{align*}& \ V(t) = 875 - 50t && \text{Let } t = 0.\\ & \ V(0) = 875 - 50(0)\\ & \ V(0) = 875 -0\\ & \boxed{V(0) = \875}\end{align*}

The original cost of the camera was 875. Vocabulary Function A function is a set of ordered pairs \begin{align*}(x, y)\end{align*} that shows a relationship where there is only one output for every input. In other words, for every value of \begin{align*}x\end{align*}, there is only one value for \begin{align*}y\end{align*}. Guided Practice 1. If \begin{align*}f(x)=3x^2-4x+6\end{align*} find: i) \begin{align*}f(-3)\end{align*} ii) \begin{align*}f(a-2)\end{align*} 2. If \begin{align*}f(m)=\frac{m+3}{2m-5}\end{align*} find ‘\begin{align*}m\end{align*}’ if \begin{align*}f(m) = \frac{12}{13}\end{align*} 3. The emergency brake cable in a truck parked on a steep hill breaks and the truck rolls down the hill. The distance in feet, \begin{align*}d\end{align*}, which the truck rolls, is represented by the function \begin{align*}d = f(t)=0.5t^2\end{align*}. i) How far will the truck roll after 9 seconds? ii) How long will it take the truck to hit a tree which is at the bottom of the hill 600 feet away? Round your answer to the nearest second. Answers 1. \begin{align*}f(x) = 3x^2 - 4x + 6\end{align*} i) \begin{align*}& \quad f(x) = 3x^2-4x+6 && \text{Substitute }(-3) \text{ for } x' \text{ in the function.}\\ & \ f({\color{red}-3}) = 3({\color{red} -3})^2 -4({\color{red}-3})+6 && \text{Perform the indicated operations.}\\ & \ f(-3) = 3({\color{red}9}) + 12 + 6 && \text{Simplify}\\ & \ f(-3) = 27 + 12 + 6\\ & \ f(-3) = {\color{red}45}\\ & \boxed{f(-3) = 45} \end{align*} ii) \begin{align*}& \qquad f(x) = 3x^2 - 4x +6\\ & \ f({\color{red}a-2}) = 3({\color{red}a-2})^2 -4 ({\color{red}a-2}) + 6 && \text{Write } (a-2)^2 \text{ in expanded form.}\\ & \ f({\color{red}a-2}) = 3({\color{red}a-2})({\color{red}a-2}) - 4({\color{red}a-2})+6 && \text{Perform the indicated operations.}\\ & \ f({\color{red}a-2}) = ({\color{red}3a-6})({\color{red}a-2}) - 4({\color{red}a-2})+6\\ & \ f(a-2) = {\color{red}3a^2-6a-6a+12-4a+8}+6 && \text{Simplify}\\ & \ f(a-2) = {\color{red}3a^2-16a+26}\\ & \boxed{f(a-2) = 3a^2-16a+26}\end{align*} 2. \begin{align*}& \qquad \qquad \ \ f(m) = \frac{m+3}{2m-5}\\ & \qquad \qquad \quad \ \ {\color{red}\frac{12}{13}} = \frac{m+3}{2m-5} && \text{Solve the equation for } m'.\\ & {\color{red}(13)(2m-5)} \frac{12}{13} = {\color{red}(13)(2m-5)} \frac{m+3}{2m-5}\\ & {\color{red}\cancel{(13)} (2m-5)} \frac{12}{\cancel{13}} = {\color{red}(13)\cancel{(2m-5)}} \frac{m+3}{\cancel{2m-5}}\\ & \qquad {\color{red}(2m-5)} 12 = {\color{red}(13)} m+3\\ & \qquad \ \ 24m-60 = 13m+39\\ & \ \ 24m-60 {\color{red}+60} = 13m + 39 {\color{red}+60}\\ & \qquad \qquad \ \ 24m = 13m+99\\ & \quad \quad 24m {\color{red}-13m} = 13m {\color{red}-13m} + 99\\ & \qquad \qquad \ \ 11m = 99\\ & \qquad \qquad \ \frac{11m}{{\color{red}11}} = \frac{99}{{\color{red}11}}\\ & \qquad \qquad \ \frac{\cancel{11}m}{{\color{red}\cancel{11}}} = \frac{\overset{9}{\cancel{99}}}{{\color{red}\cancel{11}}}\\ & \qquad \qquad \quad \boxed{m=9}\end{align*} 3. \begin{align*}d=f(t)=0.5^2\end{align*} i) \begin{align*}& \quad \ \ d =f(t)=0.5^2 && \text{Substitute 9 for } t'.\\ & \ f({\color{red}9}) = 0.5 ({\color{red}9})^2 && \text{Perform the indicated operations.}\\ & \ f(9) = 0.5 ({\color{red}81})\\ & \boxed{f(9)=40.5 \ feet}\end{align*} After 9 seconds, the truck will roll 40.5 feet. ii) \begin{align*}& d= f(t) = 0.5t^2 && \text{Substitute 600 for } d'.\\ & \qquad {\color{red}600} = 0.5t^2 && \text{Solve for } t'.\\ & \quad \ \ \frac{600}{{\color{red}0.5}} = \frac{0.5t^2}{{\color{red}0.5}}\\ & \quad \ \ \frac{\overset{{\color{red}1200}}{\cancel{600}}}{{\color{red}\cancel{0.5}}} = \frac{\cancel{0.5}t^2}{{\color{red}\cancel{0.5}}}\\ & \quad \ 1200 = t^2\\ & \ \sqrt{{\color{red}1200}} = \sqrt{{\color{red}t^2}}\\ & \boxed{34.64 \ seconds \approx t}\end{align*} The truck will hit the tree in approximately 35 seconds. Summary Using function notation to represent a function is similar to expressing the function as a formula. The formula is used to determine the value of a function for a given value. The formula can also be used to calculate the value of the variable when a value is given for the entire function. These concepts can then be applied to real life situations to determine the value of an indicated variable. Problem Set 1. If \begin{align*}g(x)=4x^2-3x+2\end{align*}, find expressions for the following: 1. \begin{align*}g(a)\end{align*} 2. \begin{align*}g(a-1)\end{align*} 3. \begin{align*}g(a+2)\end{align*} 4. \begin{align*}g(2a)\end{align*} 5. \begin{align*}g(-a)\end{align*} 2. If \begin{align*}f(y) = 5y-3\end{align*}, determine the value of ‘\begin{align*}y\end{align*}’ when: 1. \begin{align*}f(y) = 7\end{align*} 2. \begin{align*}f(y) = -1\end{align*} 3. \begin{align*}f(y) = -3\end{align*} 4. \begin{align*}f(y) = 6\end{align*} 5. \begin{align*}f(y) = -8\end{align*} Solve the following problem: The value of a Bobby Orr rookie card \begin{align*}n\end{align*} years after its purchase is \begin{align*}V(n)=520+28n\end{align*} i) Determine the value of \begin{align*}V(6)\end{align*} and explain what the solution means. ii) Determine the value of \begin{align*}n\end{align*} when \begin{align*}V(n)=744\end{align*} and explain what this represents. iii) Determine the original price of the card. Answers 1. If \begin{align*}g(x)=4x^2-3x+2\end{align*}, find... (a) \begin{align*}g(a)\end{align*} \begin{align*}& \ g(x) = 4x^2 - 3x +2\\ & \ g({\color{red}a}) = 4({\color{red}a})^2 - 3{\color{red}a} + 2\\ & \ g({\color{red}a}) = 4{\color{red}a}^2 - 3{\color{red}a} + 2\\ & \boxed{g(a) = 4a^2 - 3a + 2}\end{align*} (c) \begin{align*}g(a + 2)\end{align*} \begin{align*}& \qquad g(x) = 4x^2-3x+2\\ & \ g({\color{red}a+2}) = 4({\color{red}a+2})^2 - 3({\color{red}a+2}) + 2\\ & \ g({\color{red}a+2}) = 4({\color{red}a+2}) ({\color{red}a+2}) - 3 ({\color{red}a+2})+2\\ & \ g({\color{red}a+2}) = ({\color{red}4a+8}) ({\color{red}a+2}) - 3 ({\color{red}a+2})+2\\ & \ g({\color{red}a+2}) = {\color{red}4a^2+8a+8a+16-3a-6} + 2\\ & \ g({\color{red}a+2}) = {\color{red}4a^2+13a+12}\\ & \boxed{g(a+2) = 4a^2 +13a + 12}\end{align*} (e) \begin{align*}g(-a)\end{align*} \begin{align*}& \quad g(x) = 4x^2 - 3x+2\\ & \ g({\color{red}-a}) = 4({\color{red}-a})^2 - 3({\color{red}-a})+2\\ & \ g({\color{red}-a}) = 4{\color{red}a}^2 {\color{red}+} 3{\color{red}a} + 2\\ & \boxed{g(-a) = 4a^2 + 3a + 2}\end{align*} 2. If \begin{align*}f(y)=5y-3\end{align*} determine... (a) \begin{align*}f(y) = 7\end{align*} \begin{align*}& f(y) = 5y - 3\\ & \quad \ {\color{red}7} = 5y - 3\\ & 7 {\color{red}+3} = 5y - 3 {\color{red}+3}\\ & \ \ {\color{red}10} = 5y\\ & \ \frac{10}{{\color{red}5}} = \frac{5y}{{\color{red}5}}\\ & \ \frac{\overset{2}{\cancel{10}}}{{\color{red}\cancel{5}}} = \frac{\cancel{5}y}{{\color{red}\cancel{5}}}\\ & \quad {\color{red}2} = y\\ & \ \ \boxed{2 = y}\end{align*} (c) \begin{align*}f(y) = -3\end{align*} \begin{align*}& \quad f(y) = 5y - 3\\ & \quad \ \ {\color{red}-3} = 5y - 3\\ & -3 {\color{red}+3} = 5y - 3 {\color{red}+3}\\ & \qquad \ {\color{red}0} = 5y\\ & \qquad \frac{0}{{\color{red}5}} = \frac{5y}{{\color{red}5}}\\ & \qquad \frac{\overset{0}{\cancel{0}}}{{\color{red}\cancel{5}}} = \frac{\cancel{5}y}{{\color{red}\cancel{5}}}\\ & \qquad \ {\color{red}0} = y\\ & \qquad \boxed{0 = y}\end{align*} (e) \begin{align*}f(y) = -8\end{align*} \begin{align*}& \quad f(y) = 5y-3\\ & \quad \ \ {\color{red}-8} = 5y - 3\\ & -8 {\color{red}+3} = 5y -3 {\color{red}+3}\\ & \quad \ \ {\color{red}-5} = 5y\\ & \quad \ \frac{-5}{{\color{red}5}} = \frac{5y}{{\color{red}5}}\\ & \quad \ \frac{\overset{-1}{\cancel{-5}}}{{\color{red}\cancel{5}}} = \frac{\cancel{5}y}{{\color{red}\cancel{5}}}\\ & \quad \ \ {\color{red}-1} = y\\ & \quad \ \boxed{-1 = y}\end{align*} Solve the following... i) \begin{align*}& \ V(n) = 520 + 28n\\ & \ V({\color{red}6}) = 520 + 28 ({\color{red}6})\\ & \ V({\color{red}6}) = 520 + {\color{red}168}\\ & \ V({\color{red}6}) = {\color{red}688}\\ & \boxed{V({\color{red}6}) = 688}\end{align*} The solution represents the value of the card six years after it was purchased. The value of the card is688.00.

ii) \begin{align*}V(n) = 520 + 28n\end{align*}

\begin{align*}& \quad \ \ V(n) = 520 + 28n\\ & \qquad \ {\color{red}744} = 520 + 28n\\ & 744 {\color{red}-520} = 520 {\color{red}-520} + 28n\\ & \qquad \ {\color{red}224} = 28n\\ & \qquad \frac{224}{{\color{red}28}} = \frac{28n}{{\color{red}28}}\\ & \qquad \frac{\overset{8}{\cancel{224}}}{{\color{red}\cancel{28}}} = \frac{\cancel{28}n}{{\color{red}\cancel{28}}}\\ & \qquad \quad {\color{red}8} = n\\ & \qquad \ \ \boxed{8 = n}\end{align*}

\begin{align*}v(n) = 744\end{align*}. This represents the value of the card 8 years after it was purchased.

iii) \begin{align*}V(n) = 520 + 28n\end{align*}

\begin{align*}& \ V(n) = 520 + 28n\\ & \ V({\color{red}0}) = 250 + 28 ({\color{red}0})\\ & \ V({\color{red}0}) = 520 + {\color{red}0}\\ & \ V({\color{red}0}) = {\color{red}520}\\ & \boxed{V(0) = 520}\end{align*}

The original purchase price of the card was \$520.00.

## Summary

In this chapter you learned about linear functions. You learned what is meant by a function, function notation and how to identify a function from a graph. You also learned how to recognize whether or not a graph represented a function or a relation by doing the vertical line test. Using function notation as a formula, you learned to determine the value for an indicated variable and how to evaluate a function for a given value.

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

Show Hide Details
Description
Tags:
Subjects: