6.3: Exercises
- Page ID
- 48984
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Find \(f+g\), \(f-g\), \(f\cdot g\) for the functions below. State their domain.
- \(f(x)=x^2+6x\), \(\quad \quad \) and \(g(x)=3x-5\)
- \(f(x)=x^3+5\), \(\quad \quad \) and \(g(x)=5x^2+7\)
- \(f(x)=3x+7\sqrt{x}\), \(\quad \) and \(g(x)=2x^2+5\sqrt{x}\)
- \(f(x)=\dfrac{1}{x+2}\), \(\quad \) and \(g(x)=\dfrac{5x}{x+2}\)
- \(f(x)=\sqrt{x-3}\), \(\quad \quad \) and \(g(x)=2\sqrt{x-3}\)
- \(f(x)=x^2+2x+5\), \(\quad \quad \) and \(g(x)=3x-6\)
- \(f(x)=x^2+3x\), \(\quad \quad \) and \(g(x)=2x^2+3x+4\)
- Answer
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- \((f+g)(x)=x^{2}+9 x-5\) with domain \(D_{f+g}=\mathbb{R},(f-g)(x)= x^2 + 3x + 5\) with domain \(D_{f-g}=\mathbb{R},(f \cdot g)(x)=3 x^{3}+13 x^{2}-30 x\) with domain \(D_{f \cdot g}=\mathbb{R}\)
- \((f+g)(x)=x^{3}+5 x^{2}+12, D_{f+g}=\mathbb{R},(f-g)(x)=x^{3}-5 x^{2}-2, D_{f-g}=\mathbb{R},(f \cdot g)(x)=5 x^{5}+7 x^{3}+25 x^{2}+35, D_{f \cdot g}=\mathbb{R}\)
- \((f+g)(x)=2 x^{2}+3 x+12 \sqrt{x}, D_{f+g}=[0, \infty),(f-g)(x)=-2 x^{2}+3 x+2 \sqrt{x}, D_{f-g}=[0, \infty),(f \cdot g)(x)=6 x^{3}+14 x^{2} \sqrt{x}+15 x \sqrt{x}+35 x, D_{f \cdot g}=[0, \infty)\)
- \((f+g)(x)=\dfrac{5 x+1}{x+2}, D_{f+g}=\mathbb{R}-\{-2\},(f-g)(x)=\dfrac{1-5 x}{x+2}, D_{f-g}=\mathbb{R}-\{-2\},(f \cdot g)(x)=\dfrac{5 x}{(x+2)^{2}}, D_{f \cdot g}=\mathbb{R}-\{-2\}\)
- \((f+g)(x)=3 \sqrt{x-3}, D_{f+g}=[3, \infty),(f-g)(x)=-\sqrt{x-3}, D_{f-g}=[3, \infty),(f \cdot g)(x)=2 \cdot(\sqrt{x-3})^{2}=2 \cdot(x-3), D_{f \cdot g}=[3, \infty)\)
- \((f+g)(x)=x^{2}+5 x-1, D_{f+g}=\mathbb{R},(f-g)(x)=x^{2}-x+11, D_{f-g}=\mathbb{R},(f \cdot g)(x)=3 x^{3}+3 x-30, D_{f \cdot g}=\mathbb{R}\)
- \((f+g)(x)=3 x^{2}+6 x+4, D_{f+g}=\mathbb{R},(f-g)(x)=-x^{2}-4, D_{f-g}=\mathbb{R},(f \cdot g)(x)=2 x^{4}+9 x^{3}+13 x^{2}+12 x, D_{f \cdot g}=\mathbb{R}\)
Find \(\dfrac f g\), and \(\dfrac g f\) for the functions below. State their domain.
- \(f(x)=3x+6\), \(\quad \quad \) and \(g(x)=2x-8\)
- \(f(x)=x+2\), \(\quad \quad \) and \(g(x)=x^2-5x+4\)
- \(f(x)=\dfrac{1}{x-5}\), \(\quad \quad \) and \(g(x)=\dfrac{x-2}{x+3}\)
- \(f(x)=\sqrt{x+6}\), \(\quad \quad \) and \(g(x)=2x+5\)
- \(f(x)=x^2+8x-33\), \(\quad \quad \) and \(g(x)=\sqrt{x}\)
- Answer
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- \(\left(\dfrac{f}{g}\right)(x)=\dfrac{3 x+6}{2 x-8}\) with domain \(D_{\frac{f}{g}}=\mathbb{R}-\{4\}\), \(\left(\dfrac{g}{f}\right)(x)=\dfrac{2 x-8}{3 x+6}\) with domain \(D_{\frac{g}{f}}=\mathbb{R}-\{-2\}\)
- \(\left(\dfrac{f}{g}\right)(x)=\dfrac{x+2}{x^{2}-5 x+4}=\dfrac{x+2}{(x-4)(x-1)}\), \(D_{\frac{f}{g}}=\mathbb{R}-\{1,4\}, \quad\left(\dfrac{g}{f}\right)(x)=\dfrac{x^{2}-5 x+4}{x+2}, D_{\frac{g}{f}}=\mathbb{R}-\{-2\}\)
- \(\left(\dfrac{f}{g}\right)(x)=\dfrac{x+3}{(x-5)(x-2)}, D_{\frac{f}{g}}=\mathbb{R}-\{-3,2,5\},\left(\dfrac{g}{f}\right)(x)=\dfrac{(x-5)(x-2)}{x+3}, D_{\frac{g}{f}}=\mathbb{R}-\{-3,5\}\)
- \(\left(\dfrac{f}{g}\right)(x)=\dfrac{\sqrt{x+6}}{2 x+5}, D_{\frac{f}{g}}=\left[-6,-\dfrac{5}{2}\right) \cup\left(-\dfrac{5}{2}, \infty\right), \left(\dfrac{g}{f}\right)(x)=\dfrac{2 x+5}{\sqrt{x+6}}, D_{\frac{g}{f}}=(-6, \infty)\)
- \(\left(\dfrac{f}{g}\right)(x)=\dfrac{x^{2}+8 x-33}{\sqrt{x}}, D_{\frac{f}{g}}=(0, \infty), \left(\dfrac{g}{f}\right)(x)=\dfrac{\sqrt{x}}{x^{2}+8 x-33}, D_{\frac{g}{f}}=[0,3) \cup(3, \infty)\)
Let \(f(x)=2x-3\) and \(g(x)=3x^2+4x\). Find the following compositions
- \(f(g(2))\)
- \(g(f(2))\)
- \(f(f(5))\)
- \(f(5 g(-3))\)
- \(g(f(2)-2)\)
- \(f(f(3)+g(3))\)
- \(g(f(2+x))\)
- \(f(f(-x))\)
- \(f( f(-3)-3 g(2))\)
- \(f(f(f(2)))\)
- \(f(x+h)\)
- \(g(x+h)\)
- Answer
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- \(37\)
- \(7\)
- \(11\)
- \(147\)
- \(-1\)
- \(81\)
- \(12 x^{2}+20 x+7\)
- \(-4 x-9\)
- \(-141\)
- \(-5\)
- \(2 x+2 h-3\)
- \(3 x^{2}+6 x h+3 h^{2}+4 x+4 h\)
Find the composition \((f\circ g)(x)\) for the functions:
- \(f(x)=3x-5\), \(\quad \quad \) and \(g(x)=2x+3\)
- \(f(x)=x^2+2\), \(\quad \quad \) and \(g(x)=x+3\)
- \(f(x)=x^2-3x+2\), \(\quad \quad \) and \(g(x)=2x+1\)
- \(f(x)=x^2+\sqrt{x+3}\), \(\quad \quad \) and \(g(x)=x^2+2x\)
- \(f(x)=\dfrac{2}{x+4}\), \(\quad \quad \) and \(g(x)=x+h\)
- \(f(x)=x^2+4x+3\), \(\quad \quad \) and \(g(x)=x+h\)
- Answer
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- \((f \circ g)(x)=6 x+4\)
- \((f \circ g)(x)=x^{2}+6 x+11\)
- \((f \circ g)(x)=4 x^{2}-2x\)
- \((f \circ g)(x)=x^{4}+4 x^{3}+4 x^{2}+\sqrt{x^{2}+2 x+3}\)
- \((f \circ g)(x)=\dfrac{2}{x+h+4}\)
- \((f \circ g)(x)=x^{2}+2 x h+h^{2}+4 x+4 h+3\)
Find the compositions
\[(f\circ g)(x),\quad (g\circ f)(x),\quad(f\circ f)(x),\quad(g\circ g)(x) \nonumber \]
for the following functions:
- \(f(x)=2x+4\), \(\quad \quad \) and \(g(x)=x-5\)
- \(f(x)=x+3\), \(\quad \quad \) and \(g(x)=x^2-2x\)
- \(f(x)=2x^2-x-6\), \(\quad \quad \) and \(g(x)=\sqrt{3x+2}\)
- \(f(x)=\dfrac{1}{x+3}\), \(\quad \quad \) and \(g(x)=\dfrac{1}{x}-3\)
- \(f(x)=(2x-7)^2\), \(\quad \quad \) and \(g(x)=\dfrac{\sqrt{x}+7}{2}\)
- Answer
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- \((f \circ g)(x)=2 x-6, (g \circ f)(x)=2 x-1, (f \circ f)(x)=4 x+12, (g \circ g)(x)=x-10\)
- \((f \circ g)(x)=x^{2}-2 x+3, (g \circ f)(x)=x^{2}+4 x+3, (f \circ f)(x)=x+6, (g \circ g)(x)=x^{4}-4 x^{3}+2 x^{2}+4 x\)
- \((f \circ g)(x)=6 x-2-\sqrt{3 x+2}, (g \circ f)(x)=\sqrt{6 x^{2}-3 x-16}, (f \circ f)(x)=8 x^{4}-8 x^{3}-48 x^{2}+25 x+72, (g \circ g)(x)=\sqrt{3 \sqrt{3 x+2}+2}\)
- \((f \circ g)(x)=x, (g \circ f)(x)=x, (f \circ f)(x)=\dfrac{x+3}{3 x+10}, (g \circ g)(x)=\dfrac{10 x-3}{1-3 x}\)
- \((f \circ g)(x)=x,(g \circ f)(x)=x,(f \circ f)(x)=\left(2(2 x-7)^{2}-7\right)^{2}\) or expanded in descending degrees: \((f \circ f)(x)=64 x^{4}-896 x^{3}+4592 x^{2}-10192 x+8281,(g \circ g)(x)=\dfrac{\sqrt{\frac{\sqrt{x}+7}{2}}+7}{2}=\dfrac{14+\sqrt{14+2 \sqrt{x}}}{4}\)
Let \(f\) and \(g\) be the functions defined by the following table. Complete the table given below.
\[\begin{array}{|c||c|c|c|c|c|c|c|}
\hline x & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline \hline f(x) & 4 & 5 & 7 & 0 & -2 & 6 & 4 \\
\hline g(x) & 6 & -8 & 5 & 2 & 9 & 11 & 2 \\
\hline f(x)+3 & & & & & & & \\
\hline 4 g(x)+5 & & & & & & & \\
\hline g(x)-2 f(x) & & & & & & & \\
\hline f(x+3) & & & & & & & \\
\hline
\end{array} \nonumber \]
- Answer
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\(\begin{array}{|c||c|c|c|c|c|c|c|}
\hline x & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline \hline f(x) & 4 & 5 & 7 & 0 & -2 & 6 & 4 \\
\hline g(x) & 6 & -8 & 5 & 2 & 9 & 11 & 2 \\
\hline f(x)+3 & 7 & 8 & 10 & 3 & 1 & 9 & 7 \\
\hline 4 g(x)+5 & 29 & -28 & 25 & 13 & 41 & 49 & 13 \\
\hline g(x)-2 f(x) & -2 & -18 & -9 & 2 & 13 & -1 & -6 \\
\hline f(x+3) & 0 & -2 & 6 & 4 & \text { undef. } & \text { undef. } & \text { undef. } \\
\hline
\end{array} \nonumber \)Note, however, that the complete table for \(y = f(x + 3)\) is given by:
\(\begin{array}{|c||c|c|c|c|c|c|c|}
\hline x & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\
\hline \hline f(x+3) & 4 & 5 & 7 & 0 & -2 & 6 & 4 \\
\hline
\end{array} \nonumber \)
Let \(f\) and \(g\) be the functions defined by the following table. Complete the table by composing the given functions.
\[\begin{array}{|c||c|c|c|c|c|c|}
\hline x & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline \hline f(x) & 3 & 1 & 2 & 5 & 6 & 3 \\
\hline g(x) & 5 & 2 & 6 & 1 & 2 & 4 \\
\hline(g \circ f)(x) & & & & & & \\
\hline(f \circ g)(x) & & & & & & \\
\hline(f \circ f)(x) & & & & & & \\
\hline(g \circ g)(x) & & & & & & \\
\hline
\end{array} \nonumber \]
- Answer
-
\(\begin{array}{|c||c|c|c|c|c|c|}
\hline x & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline \hline f(x) & 3 & 1 & 2 & 5 & 6 & 3 \\
\hline g(x) & 5 & 2 & 6 & 1 & 2 & 4 \\
\hline(g \circ f)(x) & 6 & 5 & 2 & 2 & 4 & 6 \\
\hline(f \circ g)(x) & 6 & 1 & 3 & 3 & 1 & 5 \\
\hline(f \circ f)(x) & 2 & 3 & 1 & 6 & 3 & 2 \\
\hline(g \circ g)(x) & 2 & 2 & 4 & 5 & 2 & 1 \\
\hline
\end{array} \nonumber \)
Let \(f\) and \(g\) be the functions defined by the following table. Complete the table by composing the given functions.
\[\begin{array}{|c||c|c|c|c|c|c|c|}
\hline x & 0 & 2 & 4 & 6 & 8 & 10 & 12 \\
\hline \hline f(x) & 4 & 8 & 5 & 6 & 12 & -1 & 10 \\
\hline g(x) & 10 & 2 & 0 & -6 & 7 & 2 & 8 \\
\hline(g \circ f)(x) & & & & & & & \\
\hline(f \circ g)(x) & & & & & & & \\
\hline(f \circ f)(x) & & & & & & & \\
\hline(g \circ g)(x) & & & & & & & \\
\hline
\end{array} \nonumber \]
- Answer
-
\(\begin{array}{|c||c|c|c|c|c|c|c|}
\hline x & 0 & 2 & 4 & 6 & 8 & 10 & 12 \\
\hline \hline f(x) & 4 & 8 & 5 & 6 & 12 & -1 & 10 \\
\hline g(x) & 10 & 2 & 0 & -6 & 7 & 2 & 8 \\
\hline(g \circ f)(x) & 0 & 7 & \text { undef. } & -6 & 8 & \text { undef. } & 2 \\
\hline(f \circ g)(x) & -1 & 8 & 4 & \text { undef. } & \text { undef. } & 8 & 12 \\
\hline(f \circ f)(x) & 5 & 12 & \text { undef. } & 6 & 10 & \text { undef. } & -1 \\
\hline(g \circ g)(x) & 2 & 2 & 10 & \text { undef. } & \text { undef. } & 2 & 7 \\
\hline
\end{array} \nonumber \)