<meta http-equiv="refresh" content="1; url=/nojavascript/"> Function Notation | CK-12 Foundation
Dismiss
Skip Navigation
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I - Honors Go to the latest version.

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

Khan Academy Functions as Graphs

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.

Image Attributions

Description

Grades:

Date Created:

Jan 16, 2013

Last Modified:

Jun 04, 2014
Files can only be attached to the latest version of None

Reviews

Please wait...
Please wait...
Image Detail
Sizes: Medium | Original
 
CK.MAT.ENG.SE.1.Algebra-I---Honors.3.2

Original text