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Exam 2: Functions

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Question 31

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Find the inverse function of $f ( x ) = 3 ( x + 1 ) ^ { 2 }$ , $x < - 1$ .

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, but...

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Question 32

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Let $f ( x ) = - \frac { | x | } { x ^ { 2 } }$ . (a) Sketch the graph of $f$ . (b) Find the domain and the range of $f$ .

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Question 33

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Sketch the graph of the piecewise-defined function. $f ( x ) = \left\{ \begin{array} { l l } x ^ { 2 } & \text { if } | x | \leq 1 \\- x ^ { 2 } & \text { if } | x | > 1\end{array} \right.$

Determine whether or not the function $h ( x ) = x ^ { 3 } - x$ is one-to-one.

Let $f ( x ) = 3 + x ^ { 2 }$ , $0 \leq x \leq 5$ (a) Sketch the graph of $f$ then use it to sketch the graph of $f ^ { - 1 }$ . (b) Find $f ^ { - 1 }$ .

Determine whether the given curve is the graph of a function of $x$ . If it is, state the domain and range of the function.

Find the inverse function of $f ( x ) = 2 - \frac { 1 } { x }$ , $x \neq 0$ .

If $f ( x ) = \sqrt { 2 - 2 x }$ and $g ( x ) = \sqrt { x ^ { 2 } - 1 }$ , find $f + g$ , $\mathrm { fg }$ , and their domains.

A function is given. (a) Find all the local maximum and minimum values of the function and the value of x at which each occurs. (b) Find the intervals on which the function is increasing and on which the function is decreasing. State all answers correct to two decimal places. $G ( x ) = \frac { 2 } { x ^ { 2 } + x + 1 }$

If $f ( x ) = 2 x ^ { 2 } + x$ and $g ( x ) = 3 x - 1$ , find $f + g$ , $f - g$ , and their domains.

Given $f ( x ) = \frac { 1 } { x }$ , $g ( x ) = \frac { x } { x + 1 }$ , and $h ( x ) = \frac { x + 1 } { x }$ , find $f \circ g \circ h$ .

A man is running around a circular track that is 200 m in circumference. An observer uses a stopwatch to record the runner's time at the end of each lap, obtaining the data in the following table. What was the man's average speed (rate) between 108 s and 203 s? Round the answer to two decimal places. $\begin{array} { | c | c | } \hline \text { Time (s) } & \begin{array} { c } \text { Distance } \\( \mathrm { m } )\end{array} \\\hline 32 & 200 \\\hline 68 & 400 \\\hline 108 & 600 \\\hline 152 & 800 \\\hline 203 & 1000 \\\hline 263 & 1200 \\\hline 335 & 1400 \\\hline 412 & 1600 \\\hline\end{array}$

Graphs of the functions f and g are given. (a) Which is larger, $f ( - 3 ) \text { or } g ( 3 ) ?$ (b) Which is larger, $f ( - 1 ) \text { or } g ( - 1 ) ?$ (c) For which values of x is $f ( x ) = g ( x ) ?$

For the functions $f ( x ) = \frac { 1 } { x - 2 }$ and $g ( x ) = \frac { x } { x - 1 }$ , find $f + g$ , $f - g$ , $\mathrm { fg }$ , $\frac { f } { g }$ and their domains. Answer $f + g = \frac { 1 } { x - 2 } + \frac { x } { x - 1 } = \frac { x ^ { 2 } - x - 1 } { ( x - 2 ) ( x - 1 ) } \quad$ domain: $( - \infty , 1 ) \cup ( 1,2 ) \cup ( 2 , \infty )$ $\begin{array} { l l } f - g = \frac { 1 } { x - 2 } - \frac { x } { x - 1 } = - \frac { x ^ { 2 } - 3 x + 1 } { ( x - 2 ) ( x - 1 ) } & \text { domain: } ( - \infty , 1 ) \cup ( 1,2 ) \cup ( 2 , \infty ) \\f g = \left( \frac { 1 } { x - 2 } \right) \left( \frac { x } { x - 1 } \right) = \frac { x } { ( x - 1 ) ( x - 2 ) } & \text { domain: } ( - \infty , 1 ) \cup ( 1,2 ) \cup ( 2 , \infty ) \\\frac { f } { g } = \frac { ( x - 1 ) } { ( x - 2 ) x } & \text { domain: } ( - \infty , 0 ) \cup ( 0,1 ) \cup ( 1,2 ) \cup ( 2 , \infty )\end{array}$ 10. Given $f ( x ) = 2 x - 2$ and $g ( x ) = 2 x ^ { 2 }$ , find $( g \circ f ) ( - 1 )$ .

For the function $f ( x ) = 2 x ^ { 2 } - 1$ , find $f ( x + 1 )$ and $f ( x ) + 1$ .

Let $f ( x ) = 2 x - 1$ . (a) Sketch the graph of $f$ . (b) Find the domain of $f$ . (c) State the intervals on which $f$ is increasing and on which $f$ is decreasing.

Let $f ( x ) = 3 x + 1$ and $g ( x ) = 3 x ^ { 2 } - 2 x + 1$ . Find $f - g$ , $\mathrm { fg }$ , and their domains.

If $f ( x ) = 2 x ^ { 2 } + 1$ and $g ( x ) = x - 1$ , find $f + g$ , $\mathrm { fg }$ , and their domains.

A function is given. Use a graphing calculator to draw the graph of f. Find the domain and range of f from the graph. $f ( x ) = x ^ { 2 } , \quad - 3 \leq x \leq 5$

Determine whether or not the function $g ( x ) = 1 - x ^ { 2 } , x \geq 0$ is one-to-one.