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3.3: Function Notation

Difficulty Level: Advanced Created by: CK-12
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Suppose the value \begin{align*}V\end{align*}V of a digital camera \begin{align*}t\end{align*}t years after it was bought is represented by the function \begin{align*}V(t) = 875 - 50t\end{align*}V(t)=87550t.

  • Can you determine the value of \begin{align*}V(4)\end{align*}V(4) and explain what the solution means in the context of this problem?
  • Can you determine the value of \begin{align*}t\end{align*}t then \begin{align*}V(t) = 525\end{align*}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 \begin{align*}x\end{align*}x into \begin{align*}3x - 1\end{align*}3x1. So, for example, if the function is named \begin{align*}f\end{align*}f, and 3 is fed into the machine, \begin{align*}3(3) - 1 = 8\end{align*}3(3)1=8 comes out.

When naming a function the symbol \begin{align*}f(x)\end{align*}f(x) is often used. The symbol \begin{align*}f(x)\end{align*}f(x) is pronounced as “\begin{align*}f\end{align*}f of \begin{align*}x\end{align*}x.” This means that the equation is a function that is written in terms of the variable \begin{align*}x\end{align*}x. An example of such a function is \begin{align*}f(x) = 3x+4\end{align*}f(x)=3x+4. Functions can also be written using a letter other than \begin{align*}f\end{align*}f and a variable other than \begin{align*}x\end{align*}x. For example, \begin{align*}v(t) = 2t^2 - 5\end{align*}v(t)=2t25 and \begin{align*}d(h) = 4h-3\end{align*}d(h)=4h3. 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 \begin{align*}f(x)=3x-4\end{align*}f(x)=3x4, then to calculate \begin{align*}f(4),\end{align*}f(4), substitute:

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

f(4)f(4)f(4)=3(4)4=124=8

Graphically, if \begin{align*}f(4) = 8\end{align*}f(4)=8, 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*}f(x)=x2+2x+5 find.

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

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

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

Solution:

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*}

f(x)=x2+2x+5  f(2)=(2)2+2(2)+5f(2)=4+4+5f(2)=13f(x)=x2+2x+5     f(7)=(7)2+2(7)+5 f(7)=4914+5f(7)=40f(x)=x2+2x+5     f(1.4)=(1.4)2+2(1.4)+5 f(1.4)=1.96+2.8+5f(1.4)=9.76

Example B

Functions can also be represented as mapping rules. If \begin{align*}g(x)\rightarrow 5-2x\end{align*}g(x)52x find the following in simplest form:

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

b) \begin{align*}g(y-3)\end{align*}g(y3)

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

Solution:

\begin{align*}& \ 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}\end{align*}

 g(x)52x  g(y)52(y)g(y)52y  g(x)52x g(2y)52(2y)g(2y)54yg(x)52x    g(y3)52(y3) g(y3)52y+6g(y3)112yApply the distributive propertyCombine like terms

Example C

If \begin{align*}P(a)=\frac{2a-3}{a+2}\end{align*}P(a)=2a3a+2:

a) Evaluate

i) \begin{align*}P(0)\end{align*}P(0)
ii) \begin{align*}P(1)\end{align*}P(1)
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*}

Solution:

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*}

Concept Problem Revisited

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*}

  • Determine the value of \begin{align*}V(4)\end{align*} and explain what the solution mean to this problem.
  • Determine the value of \begin{align*}t\end{align*} then \begin{align*}V(t) = 525\end{align*} 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.

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

  • The digital camera has a value of $525, 7 years after it was purchased.

\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 original cost of the camera was $875.

\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*}

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.

Practice

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*}

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*}

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*}.

  1. Determine the value of \begin{align*}V(6)\end{align*} and explain what the solution means.
  2. Determine the value of \begin{align*}n\end{align*} when \begin{align*}V(n)=744\end{align*} and explain what this represents.
  3. Determine the original price of the card.

Vocabulary

Function

Function

A function is a relation where there is only one output for every input. In other words, for every value of x, there is only one value for y.

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Date Created:

Jan 16, 2013

Last Modified:

Feb 26, 2015
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