4.2E Exercises

Last updated
Save as PDF

Page ID: 153650

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$ \newcommand{\dsum}{\displaystyle\sum\limits} $

$ \newcommand{\dint}{\displaystyle\int\limits} $

$ \newcommand{\dlim}{\displaystyle\lim\limits} $

$ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\id}{\mathrm{id}}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\kernel}{\mathrm{null}\,}$

$ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$

$ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$

$ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$

$ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vectorC}[1]{\textbf{#1}} $

$ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $

$ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $

$ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$\newcommand{\longvect}{\overrightarrow}$

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$\newcommand{\avec}{\mathbf a}$ $\newcommand{\bvec}{\mathbf b}$ $\newcommand{\cvec}{\mathbf c}$ $\newcommand{\dvec}{\mathbf d}$ $\newcommand{\dtil}{\widetilde{\mathbf d}}$ $\newcommand{\evec}{\mathbf e}$ $\newcommand{\fvec}{\mathbf f}$ $\newcommand{\nvec}{\mathbf n}$ $\newcommand{\pvec}{\mathbf p}$ $\newcommand{\qvec}{\mathbf q}$ $\newcommand{\svec}{\mathbf s}$ $\newcommand{\tvec}{\mathbf t}$ $\newcommand{\uvec}{\mathbf u}$ $\newcommand{\vvec}{\mathbf v}$ $\newcommand{\wvec}{\mathbf w}$ $\newcommand{\xvec}{\mathbf x}$ $\newcommand{\yvec}{\mathbf y}$ $\newcommand{\zvec}{\mathbf z}$ $\newcommand{\rvec}{\mathbf r}$ $\newcommand{\mvec}{\mathbf m}$ $\newcommand{\zerovec}{\mathbf 0}$ $\newcommand{\onevec}{\mathbf 1}$ $\newcommand{\real}{\mathbb R}$ $\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$ $\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$ $\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$ $\newcommand{\laspan}[1]{\text{Span}\{#1\}}$ $\newcommand{\bcal}{\cal B}$ $\newcommand{\ccal}{\cal C}$ $\newcommand{\scal}{\cal S}$ $\newcommand{\wcal}{\cal W}$ $\newcommand{\ecal}{\cal E}$ $\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$ $\newcommand{\gray}[1]{\color{gray}{#1}}$ $\newcommand{\lgray}[1]{\color{lightgray}{#1}}$ $\newcommand{\rank}{\operatorname{rank}}$ $\newcommand{\row}{\text{Row}}$ $\newcommand{\col}{\text{Col}}$ $\renewcommand{\row}{\text{Row}}$ $\newcommand{\nul}{\text{Nul}}$ $\newcommand{\var}{\text{Var}}$ $\newcommand{\corr}{\text{corr}}$ $\newcommand{\len}[1]{\left|#1\right|}$ $\newcommand{\bbar}{\overline{\bvec}}$ $\newcommand{\bhat}{\widehat{\bvec}}$ $\newcommand{\bperp}{\bvec^\perp}$ $\newcommand{\xhat}{\widehat{\xvec}}$ $\newcommand{\vhat}{\widehat{\vvec}}$ $\newcommand{\uhat}{\widehat{\uvec}}$ $\newcommand{\what}{\widehat{\wvec}}$ $\newcommand{\Sighat}{\widehat{\Sigma}}$ $\newcommand{\lt}{<}$ $\newcommand{\gt}{>}$ $\newcommand{\amp}{&}$ $\definecolor{fillinmathshade}{gray}{0.9}$

Vertical Line Test

Tell whether the following graphs could represent functions.

$hw1.png$	$hw3.png$	$hw2.png$	$hw4.png$
1.	2.	3.	4.

Answer

Domain and Range From Graphs

Give the domain and range of the following graphed functions.

$hw2.png$	$hw8.png$	$hw6.png$	$hw7.png$
1.	2.	3.	4.

Answer

Dom: $( -\infty, \infty ) $, Range: $ [-1,1] $
Dom: $ [-3, \infty) $, Range: $ [0,\infty) $
Dom: $( -\infty, \infty ) $, Range: $ [-1, \infty) $
Dom: $( -\infty, \infty ) $, Range: $ (-\infty, \infty) $

Describing Transformations

Describe the transformation(s) done to $f(x) = x^2 $ to produce the following functions.

$ g(x) = -2x^2 $
$ g(x) = x^2 + 3 $
$ g(x) = \frac{1}{4} (x-1)^2 $
$ g(x) = (x+2)^2 - 4 $
$ g(x) = -(2x+3)^2 $

Answer

Vertical stretch by a factor of 2 and reflection across the $x$-axis.
Vertical shift up by 3.
Horizontal shift right by 1, then vertical shrink by a factor of $\frac{1}{4} $. (Or, horizontal stretch by a factor of $ 2 $, and then horizontal shift right by 1.)
Horizontal shift left by 2 and vertical shift down by 4.
Horizontal shift left by 3, then horizontal shrink by a factor of $\frac{1}{2} $, then reflect over the $x$-axis. (Or, horizontal shrink by a factor of $ \frac{1}{2} $, then a horizontal shift left by $ \frac{3}{2} $, then reflect over $x$-axis.)

Producing New Functions

Write the function $g(x)$ obtained from $f(x) = x^2$ by performing the transformations in order.

Horizontal shift right by 1 and vertical shift down by 3.
Vertical stretch by a factor of 3, then reflect across the $x$-axis, then shift vertically up by 2.
Reflect across the $y$-axis, then horizontal shift left by 2, then horizontal stretch by a factor of $3$.

Answer

1. $g(x) = (x-1)^2 - 3 $

2. $g(x) = -3x^2 + 2 $

3. $g(x)= \left( \frac{1}{3}x + 2 \right)^2 $ (Note that the first step had no effect, because $ (-x)^2 = x^2 $. This function is symmetric about the $y$-axis.)

Describing Transformations From Graphs

Describe the transformations that changed the black graph into the red graph.

$hw9.png$	$hw10.png$	$hw11.png$
1.	2.	3.

Answer

Reflect across the $y$-axis, then shift up by 1.
Shift left by 2 and down by 1.
Vertical stretch by a factor of 3. (Or horizontal shrink by a factor of $ \frac{1}{3} $. This can be seen by comparing the points $ (1,1)$ on the black graph and $ (1,3) $ on the red graph. This is a picture of $ f(x) = |x| $ and $ g(x) = |3x| = 3|x| $, so it can be seen as vertical stretch or a horizontal shrink, equivalently.

Even and Odd Functions

Determine whether the function is even, odd, or neither.

$ f(x) = x^5 - x $
$ p(x) = x^3 + 1$
$ g(x) = 2x + x^2 $
$ f(x) = x^7 $
$ h(x) = x^4 - 4x^2 $
$ f(x) = |x| $

Answer

Odd
Neither
Neither
Odd
Even
Even

Sketching Graphs of Functions

Sketch the graphs of the functions by plotting points and connecting the dots, and then use them to sketch the graphs of the transformed functions.

$ f(x) = x^2$ becoming $g(x) = x^2 + 1 $
$ f(x) = \sqrt{x}$ becoming $ g(x) = - \sqrt{x} $
$ f(x) = 3x + 1$ becoming $ g(x) = -(3x+1) $

Answer

Plot the points $ (0,0), (1,1), (-1,1), (2,4), (-2,4), (3,9),$ etc. Connect the dots smoothly. Then use that to sketch $g(x)$.
Plot the points $ (0,0), (1,1), (4,2), (9,3),$ etc. Connect the dots smoothly. Use that to sketch $g (x)$.
This is a straight line, so sketch as usual by plotting $y$-intercept and following the slope. Connect the dots with a straight line. Use that to sketch $g (x)$.

$hw12.png$	$hw13.png$	$hw14.png$
1.	2.	3.

Search

Text Color

Text Size

Margin Size

Font Type