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# 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 $x$ into $3x - 1$. If the function is $f$, and 3 is fed into the machine, $3(3) - 1 = 8$ comes out.

Watch This

Guidance

To represent functions as equations, the symbol $f(x)$ is often used. The symbol $f(x)$ is pronounced as “$f$ of $x$.” This means that the equation is a function that is written in the terms of the variable $x$. An example of such a function is $f(x) = 3x+4$. Functions can also be written using a letter other than $f$ and a variable other than $x$. An example of such a case is $v(t) = 2t^2 - 5$ and $d(h) = 4h-3$. 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 $f(x)=3x-4$, then $f(4)=?$

$f(x) &= 3x-4\\f(x) & = 3(4) - 4\\f(4) & = 12-4\\f(4) & = 8$

Graphically, if $f(4) = 8$, this means that the point (4, 8) is a point on the graph of the line.

Example A

If $f(x) = x^2 + 2x +5$ find.

a) $f(2)$

b) $f(-7)$

c) $f(1.4)$

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

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

Example B

Functions can also be represented as mapping rules. If $g(x)\rightarrow 5-2x$ find the following in simplest form:

a) $g(y)$

b) $g(y-3)$

c) $g(2y)$

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

Example C

If $P(a)=\frac{2a-3}{a+2}$

a) evaluate

i) $P(0)$

ii) $P(1)$

iii) $P \left ( -\frac{1}{2} \right )$

b) Find a value of ‘$a$’ where $P(a)$ does not exist.

c) Find $P(a-2)$ in simplest form

d) Find ‘$a$’ if $P(a)=-5$

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

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

$& \quad \ \ a+ 2 = 0\\& a+2-2=0-2\\& \qquad \quad \ \boxed{a=-2}$

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

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

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

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

Example D

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

The value $V$ of a digital camera $t$ years after it was bought is represented by the function $V(t) = 875 - 50t$

a) Determine the value of $V(4)$ and explain what the solution mean to this problem.

b) Determine the value of $t$ then $V(t) = 525$ and explain what this represents.

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

a) $& \ V(t) = 875 - 50t\\& \ V(4) = 875 - 50(4)\\& \ V(4) = 875-200\\& \boxed{V(4) = \ 675}$

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

c) $& \ V(t) = 875 - 50t && \text{Let } t = 0.\\& \ V(0) = 875 - 50(0)\\& \ V(0) = 875 -0\\& \boxed{V(0) = \875}$

The original cost of the camera was $875. Vocabulary Function A function is a set of ordered pairs $(x, y)$ that shows a relationship where there is only one output for every input. In other words, for every value of $x$, there is only one value for $y$. Guided Practice 1. If $f(x)=3x^2-4x+6$ find: i) $f(-3)$ ii) $f(a-2)$ 2. If $f(m)=\frac{m+3}{2m-5}$ find ‘$m$’ if $f(m) = \frac{12}{13}$ 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, $d$, which the truck rolls, is represented by the function $d = f(t)=0.5t^2$. 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. $f(x) = 3x^2 - 4x + 6$ i) $& \quad f(x) = 3x^2-4x+6 && \text{Substitute }(-3) \text{ for } x' \text{ in the function.}\\& \ f({\color{red}-3}) = 3({\color{red} -3})^2 -4({\color{red}-3})+6 && \text{Perform the indicated operations.}\\& \ f(-3) = 3({\color{red}9}) + 12 + 6 && \text{Simplify}\\& \ f(-3) = 27 + 12 + 6\\& \ f(-3) = {\color{red}45}\\& \boxed{f(-3) = 45}$ ii) $& \qquad f(x) = 3x^2 - 4x +6\\& \ f({\color{red}a-2}) = 3({\color{red}a-2})^2 -4 ({\color{red}a-2}) + 6 && \text{Write } (a-2)^2 \text{ in expanded form.}\\& \ f({\color{red}a-2}) = 3({\color{red}a-2})({\color{red}a-2}) - 4({\color{red}a-2})+6 && \text{Perform the indicated operations.}\\& \ f({\color{red}a-2}) = ({\color{red}3a-6})({\color{red}a-2}) - 4({\color{red}a-2})+6\\& \ f(a-2) = {\color{red}3a^2-6a-6a+12-4a+8}+6 && \text{Simplify}\\& \ f(a-2) = {\color{red}3a^2-16a+26}\\& \boxed{f(a-2) = 3a^2-16a+26}$ 2. $& \qquad \qquad \ \ f(m) = \frac{m+3}{2m-5}\\& \qquad \qquad \quad \ \ {\color{red}\frac{12}{13}} = \frac{m+3}{2m-5} && \text{Solve the equation for } m'.\\& {\color{red}(13)(2m-5)} \frac{12}{13} = {\color{red}(13)(2m-5)} \frac{m+3}{2m-5}\\& {\color{red}\cancel{(13)} (2m-5)} \frac{12}{\cancel{13}} = {\color{red}(13)\cancel{(2m-5)}} \frac{m+3}{\cancel{2m-5}}\\& \qquad {\color{red}(2m-5)} 12 = {\color{red}(13)} m+3\\& \qquad \ \ 24m-60 = 13m+39\\& \ \ 24m-60 {\color{red}+60} = 13m + 39 {\color{red}+60}\\& \qquad \qquad \ \ 24m = 13m+99\\& \quad \quad 24m {\color{red}-13m} = 13m {\color{red}-13m} + 99\\& \qquad \qquad \ \ 11m = 99\\& \qquad \qquad \ \frac{11m}{{\color{red}11}} = \frac{99}{{\color{red}11}}\\& \qquad \qquad \ \frac{\cancel{11}m}{{\color{red}\cancel{11}}} = \frac{\overset{9}{\cancel{99}}}{{\color{red}\cancel{11}}}\\& \qquad \qquad \quad \boxed{m=9}$ 3. $d=f(t)=0.5^2$ i) $& \quad \ \ d =f(t)=0.5^2 && \text{Substitute 9 for } t'.\\& \ f({\color{red}9}) = 0.5 ({\color{red}9})^2 && \text{Perform the indicated operations.}\\& \ f(9) = 0.5 ({\color{red}81})\\& \boxed{f(9)=40.5 \ feet}$ After 9 seconds, the truck will roll 40.5 feet. ii) $& d= f(t) = 0.5t^2 && \text{Substitute 600 for } d'.\\& \qquad {\color{red}600} = 0.5t^2 && \text{Solve for } t'.\\& \quad \ \ \frac{600}{{\color{red}0.5}} = \frac{0.5t^2}{{\color{red}0.5}}\\& \quad \ \ \frac{\overset{{\color{red}1200}}{\cancel{600}}}{{\color{red}\cancel{0.5}}} = \frac{\cancel{0.5}t^2}{{\color{red}\cancel{0.5}}}\\& \quad \ 1200 = t^2\\& \ \sqrt{{\color{red}1200}} = \sqrt{{\color{red}t^2}}\\& \boxed{34.64 \ seconds \approx t}$ 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 $g(x)=4x^2-3x+2$, find expressions for the following: 1. $g(a)$ 2. $g(a-1)$ 3. $g(a+2)$ 4. $g(2a)$ 5. $g(-a)$ 2. If $f(y) = 5y-3$, determine the value of ‘$y$’ when: 1. $f(y) = 7$ 2. $f(y) = -1$ 3. $f(y) = -3$ 4. $f(y) = 6$ 5. $f(y) = -8$ Solve the following problem: The value of a Bobby Orr rookie card $n$ years after its purchase is $V(n)=520+28n$ i) Determine the value of $V(6)$ and explain what the solution means. ii) Determine the value of $n$ when $V(n)=744$ and explain what this represents. iii) Determine the original price of the card. Answers 1. If $g(x)=4x^2-3x+2$, find... (a) $g(a)$ $& \ g(x) = 4x^2 - 3x +2\\& \ g({\color{red}a}) = 4({\color{red}a})^2 - 3{\color{red}a} + 2\\& \ g({\color{red}a}) = 4{\color{red}a}^2 - 3{\color{red}a} + 2\\& \boxed{g(a) = 4a^2 - 3a + 2}$ (c) $g(a + 2)$ $& \qquad g(x) = 4x^2-3x+2\\& \ g({\color{red}a+2}) = 4({\color{red}a+2})^2 - 3({\color{red}a+2}) + 2\\& \ g({\color{red}a+2}) = 4({\color{red}a+2}) ({\color{red}a+2}) - 3 ({\color{red}a+2})+2\\& \ g({\color{red}a+2}) = ({\color{red}4a+8}) ({\color{red}a+2}) - 3 ({\color{red}a+2})+2\\& \ g({\color{red}a+2}) = {\color{red}4a^2+8a+8a+16-3a-6} + 2\\& \ g({\color{red}a+2}) = {\color{red}4a^2+13a+12}\\& \boxed{g(a+2) = 4a^2 +13a + 12}$ (e) $g(-a)$ $& \quad g(x) = 4x^2 - 3x+2\\& \ g({\color{red}-a}) = 4({\color{red}-a})^2 - 3({\color{red}-a})+2\\& \ g({\color{red}-a}) = 4{\color{red}a}^2 {\color{red}+} 3{\color{red}a} + 2\\& \boxed{g(-a) = 4a^2 + 3a + 2}$ 2. If $f(y)=5y-3$ determine... (a) $f(y) = 7$ $& f(y) = 5y - 3\\& \quad \ {\color{red}7} = 5y - 3\\& 7 {\color{red}+3} = 5y - 3 {\color{red}+3}\\& \ \ {\color{red}10} = 5y\\& \ \frac{10}{{\color{red}5}} = \frac{5y}{{\color{red}5}}\\& \ \frac{\overset{2}{\cancel{10}}}{{\color{red}\cancel{5}}} = \frac{\cancel{5}y}{{\color{red}\cancel{5}}}\\& \quad {\color{red}2} = y\\& \ \ \boxed{2 = y}$ (c) $f(y) = -3$ $& \quad f(y) = 5y - 3\\& \quad \ \ {\color{red}-3} = 5y - 3\\& -3 {\color{red}+3} = 5y - 3 {\color{red}+3}\\& \qquad \ {\color{red}0} = 5y\\& \qquad \frac{0}{{\color{red}5}} = \frac{5y}{{\color{red}5}}\\& \qquad \frac{\overset{0}{\cancel{0}}}{{\color{red}\cancel{5}}} = \frac{\cancel{5}y}{{\color{red}\cancel{5}}}\\& \qquad \ {\color{red}0} = y\\& \qquad \boxed{0 = y}$ (e) $f(y) = -8$ $& \quad f(y) = 5y-3\\& \quad \ \ {\color{red}-8} = 5y - 3\\& -8 {\color{red}+3} = 5y -3 {\color{red}+3}\\& \quad \ \ {\color{red}-5} = 5y\\& \quad \ \frac{-5}{{\color{red}5}} = \frac{5y}{{\color{red}5}}\\& \quad \ \frac{\overset{-1}{\cancel{-5}}}{{\color{red}\cancel{5}}} = \frac{\cancel{5}y}{{\color{red}\cancel{5}}}\\& \quad \ \ {\color{red}-1} = y\\& \quad \ \boxed{-1 = y}$ Solve the following... i) $& \ V(n) = 520 + 28n\\& \ V({\color{red}6}) = 520 + 28 ({\color{red}6})\\& \ V({\color{red}6}) = 520 + {\color{red}168}\\& \ V({\color{red}6}) = {\color{red}688}\\& \boxed{V({\color{red}6}) = 688}$ The solution represents the value of the card six years after it was purchased. The value of the card is$688.00.

ii) $V(n) = 520 + 28n$

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

$v(n) = 744$. This represents the value of the card 8 years after it was purchased.

iii) $V(n) = 520 + 28n$

$& \ V(n) = 520 + 28n\\& \ V({\color{red}0}) = 250 + 28 ({\color{red}0})\\& \ V({\color{red}0}) = 520 + {\color{red}0}\\& \ V({\color{red}0}) = {\color{red}520}\\& \boxed{V(0) = 520}$

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.

Jan 16, 2013

Jan 14, 2015