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3.3: Function Notation

Difficulty Level: Advanced Created by: CK-12
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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 then V(t) = 525 and explain what this represents?
  • What was the original cost of the digital camera?

Watch This

Khan Academy Functions as Graphs

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
  • 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. Suppose f(x)=3x-4 , then 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:

& \ g(x) \rightarrow 5 -2x && \qquad g(x) \rightarrow 5-2x \\& \ {\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

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) 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}

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 mean to this problem.
  • Determine the value of t then V(t) = 525 and explain what this 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 , 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.

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 represents.
  3. Determine the original price of the card.

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Difficulty Level:

Advanced

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Date Created:

Jan 16, 2013

Last Modified:

Jul 15, 2014
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