4.3E Exercises
- Page ID
- 153661
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Given each pair of functions, find \( f+g, f-g, fg,\) and \( \dfrac{f}{g} \), and give their domains.
- \( f(x) = x^4 + x^2, \: \: g(x) = x^3 + 1 \)
- \( f(x) = \sqrt{25-x^2}, \: \: g(x) = x+1\)
- \( f(x) = \dfrac{x}{x+1}, \: \: g(x) = x^2 + 2 \)
- Answer
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1.
- \( (f+g)(x) = x^4 + x^3+ x^2 + 1 \), dom: \( (-\infty, \infty) \)
- \( (f-g)(x) = x^4 -x^3 + x^2 - 1 \), dom: \( (-\infty, \infty) \)
- \( (fg)(x) = x^7 + x^5 + x^4 + x^2\), dom: \( (-\infty, \infty) \)
- \( \left( \frac{f}{g} \right) (x) = \dfrac{x^4+x^2}{x^3+1}\), dom: \( x \neq -1 \)
2.
- \( (f+g)(x) = \sqrt{25-x^2} + x + 1 \), dom: \( [-5,5] \)
- \( (f-g)(x) =\sqrt{25-x^2} - x - 1 \), dom: \( [-5,5] \)
- \( (fg)(x) = (x+1)\sqrt{25-x^2} \), dom: \( [-5,5] \)
- \( \left( \frac{f}{g} \right) (x) = \dfrac{ \sqrt{25-x^2}}{x+1} \), dom: \( \{ x \: | \: -5 \leq x \leq 5, x \neq -1 \} \) or \( [-5,-1) \cup (-1, 5] \)
3.
- \( (f+g)(x) = \dfrac{x}{x+1}+ x^2 +2 \), dom: \( x \neq -1 \)
- \( (f-g)(x) = \dfrac{x}{x+1} - x^2 - 2\), dom: \( x \neq -1 \)
- \( (fg)(x) = \dfrac{x(x^2+2)}{x+1} \), dom: \( x \neq -1 \)
- \( \left( \frac{f}{g} \right) (x) = \dfrac{x}{(x+1)(x^2+2)} \), dom: \(x \neq - 1 \)
Given each pair of functions, find \( f \circ g, g \circ f, f \circ f,\) and \( g \circ g \). Simplify if reasonable.
- \( f(x) = x^4 + x^2, \: \: g(x) = x^2 + 1 \)
- \( f(x) = \sqrt{25-x^2}, \: \: g(x) = x+1\)
- \( f(x) = \dfrac{x}{x+1}, \: \: g(x) = x^2 - 1 \)
- Answer
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1.
- \( (f \circ g)(x) = (x^2+1)^4 + (x^2+1)^2 \)
- \( (g \circ f)(x) = (x^4 + x^2)^2 + 1 \)
- \( (f\circ f)(x) = (x^4 + x^2)^4 + (x^4 + x^2)^2 \)
- \( (g \circ g)(x) = (x^2+1)^2 + 1 = x^4+2x^2 + 2 \)
2.
- \( (f \circ g)(x) =\sqrt{25 - (x+1)^2} \)
- \( (g \circ f)(x) =\sqrt{25-x^2} + 1 \)
- \( (f\circ f)(x) = \sqrt{25 - (\sqrt{25-x^2})^2} = \sqrt{25 - 25 + x^2} = \sqrt{x^2} = |x| \)
- \( (g \circ g)(x) =x+2 \)
3.
- \( (f \circ g)(x) =\dfrac{x^2-1}{x^2} = 1 - \dfrac{1}{x^2} \)
- \( (g \circ f)(x) =\left( \dfrac{x}{x+1} \right)^2 - 1 \)
- \( (f\circ f)(x) = \dfrac{ \frac{x}{x+1}} {\frac{x}{x+1}+1} = ... = \dfrac{x}{2x+1} \) (Hint: get a common denominator in the big denominator. )
- \( (g \circ g)(x) =(x^2-1)^2 - 1 = x^4 - 2x^2 \)
Given three functions \(f, g,\) and \(h\), find \( h \circ g \circ f\).
- \( f(x) = x^2, \: \: g(x) = x + 4, \: \: h(x) = \sqrt{x} \)
- \( f(x) = x -1, \: \: g(x) = x^3, \: \: h(x) = \dfrac{1}{x} \)
- \( f(x) = 2, \: \: g(x) = \sqrt{x+1}, \: \: h(x) = x^2 \)
- Answer
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- \( (h\circ g\circ f)(x) = \sqrt{x^2+4} \)
- \( (h\circ g\circ f)(x) = \dfrac{1}{(x-1)^3} \)
- \( (h\circ g\circ f)(x) = (\sqrt{2+1})^2 = 3 \)
Find two functions \(f\) and \(g\) such that \(h = f \circ g\).
- \( h(x) = (x - 3)^5 \)
- \( h(x) = \dfrac{x^2}{x^2+1} \)
- \( h(x) = (x-1)^3 + (x-1)^2 + 3 \)
- Answer
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- \( g(x) = x-3, f(x) = x^5 \)
- \( g(x) = x^2, f(x) = \dfrac{x}{x+1} \)
- \( g(x) = x-1, f(x) = x^3 + x^2 + 3 \)
If I'm blowing up a spherical balloon, then the radius of the balloon is growing as time goes on. Aka, I can see the radius \(r\) as a function of time \(t\). Let's say that the radius is growing at a rate of 3 inches per second, so the function giving the radius at time \(t\) is \(r(t) = 3t\). Meanwhile, I see volume \(V\) as a function of radius \(r\) with the formula \(V(r) = \frac{4}{3} \pi r^3 \). Use composition to write volume as a function of time, \( V(t)\). (This type of idea will come up when you learn about related rates in Calc I.)
- Answer
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Volume \(V\) depends on radius \(r\), which itself depends on time \(t\). I can daisychain \( V(r) = V(r(t)) = (V \circ r)(t)\) to have a function that will take a time \(t\) as input and give a volume \( V(t)\) as output. With these formulas, we have
\[ V(r(t)) = V(3t) = \frac{4}{3} \pi (3t)^3 = 36 \pi t^3 \notag \]
Say water is pouring into a cylindrical tank such that the height of the water level is increasing by 5 inches per minute. At time \(t = 0\), the tank started out empty, and the cylinder's base has radius 15 inches. Write a function that expresses the volume of water in the tank as a function of time. (Recall the volume of a cylinder can be computed using the radius and the height, \(V = \pi r^2 h \).)
- Answer
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We have \( h(t) = 5t \) and \(V(h) = \pi (15)^2 h = 225 \pi h \). Then \(V(t) = V(h(t)) = 225\pi (5t) = 1125\pi t \).