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3.3: Function Notation

Difficulty Level: Advanced Created by: CK-12
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Practice Function Notation
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Suppose the value $V$ of a digital camera $t$ years after it was bought is represented by the function $V(t) = 875 - 50t$ .

• Can you determine the value of $V(4)$ and explain what the solution means in the context of this problem?
• Can you determine the value of $t$ when $V(t) = 525$ and explain what this situation represents?
• What was the original cost of the digital camera?

Guidance

A function machine shows how a function responds to an input. If I triple the input and subtract one, the machine will convert $x$ into $3x - 1$ . So, for example, if the function is named $f$ , and 3 is fed into the machine, $3(3) - 1 = 8$ comes out.

When naming a function 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 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 $x$ . For example, $v(t) = 2t^2 - 5$ and $d(h) = 4h-3$ . In addition to representing a function as an equation, you can also represent a function:

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

When a function is represented as an equation, an ordered pair can be determined by evaluating various values of the assigned variable. Suppose $f(x)=3x-4$ . To calculate $f(4),$ substitute:

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

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) $f(1.4)$

Solution:

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) $g(2y)$

Solution:

a) $g(y)=5-2y$

b) $g(y-3)=5-2(y-3)=5-2y+6=11-2y$

c) $g(2y)=5-2(2y)=5-4y$

Example C

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

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 $P(a)=-5$

Solution:

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) The function will not exist if the denominator equals zero because division by zero is undefined.

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

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

c)

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

d)

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

Concept Problem Revisited

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

• Determine the value of $V(4)$ and explain what the solution means to this problem.
• Determine the value of $t$ when $V(t) = 525$ and explain what this situation represents.
• What was the original cost of the digital camera?

Solution:

• The camera is valued at $675, 4 years after it was purchased. $& \ V(t) = 875 - 50t\\& \ V(4) = 875 - 50(4)\\& \ V(4) = 875-200\\& \boxed{V(4) = \ 675}$ • The digital camera has a value of$525, 7 years after it was purchased.

$& \qquad \ V(t) = 875 - 50t && \text{Let } V(t) = 525\\& \qquad \ \ 525 = 875-50t && \text{Solve the equation}\\& 525 -875 = 875 - 875 - 50t\\& \quad \ -350 = - 50t\\& \quad \ \ \frac{-350}{-50} = \frac{-50t}{-50}\\& \qquad \quad \ \boxed{7 = t}$

• The original cost of the camera was \$875.

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

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, $d$ , that 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.

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.

Practice

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. $g(-a)$

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. $f(y) = -8$

The value of a Bobby Orr rookie card $n$ years after its purchase is $V(n)=520+28n$ .

1. Determine the value of $V(6)$ and explain what the solution means.
2. Determine the value of $n$ when $V(n)=744$ and explain what this situation represents.
3. Determine the original price of the card.

Let $f(x)=\frac{3x}{x+2}$ .

1. When is $f(x)$ undefined?
2. For what value of $x$ does $f(x)=2.4$ ?

Jan 16, 2013

Jul 15, 2014