4.4E Exercises
- Page ID
- 153684
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Investigate the functions algebraically and intuitively and determine if you think they are one-to-one.
- The floor function. (see Section 4.2)
- The absolute value function.
- Any linear function \( f(x) = mx + b\).
- The square root function.
- Answer
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- No. For example, the floor function sends all real numbers \( 1 \leq x < 2\) to the same value, \(1\).
- No. The absolute value function's graph looks like a V, which would fail the HLT.
- Yes, any straight line graph will pass the HLT.
- Yes, the square root function by convention only returns the positive root.
Determine whether the graphed functions are one-to-one.
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| 1. | 2. | 3. | 4. |
- Answer
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- Yes
- Yes
- No
- No
Using the graph of \(f(x)\), find \( f^{-1}(-2), f^{-1}(-1), f^{-1}(0), f^{-1}(1),\) and \( f^{-1}(2) \). Try to identify the function that was graphed, as well as its inverse function.
- Answer
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- \( f^{-1}(-2) = -8\)
- \( f^{-1}(-1)= -1 \)
- \( f^{-1}(0) = 0 \)
- \( f^{-1}(1) = 1\)
- \( f^{-1}(2) = 8\).
The graphed function is \( f(x) = \sqrt[3]{x} \). The inverse function is \(f^{-1}(x) = x^3 \).
Test whether the functions are inverses of each other by checking the conditions:
\[ (g \circ f)(x) = x, \text{ for all } x \text{ in domain of } f, \quad \text{and} \quad (f \circ g)(x) = x, \text{ for all } x \text{ in domain of } g \notag \]
- \( f(x) = 14-3x\) and \(g(x) = \dfrac{14-x}{3} \)
- \( f(x) = (x-5)^3 \) and \(g(x) = \sqrt[3]{x} + 5 \)
- \( f(x) = x\) and \( g(x) = \frac{1}{x} \)
- \( f(x) = \dfrac{x+1}{x+2} \) and \( g(x) = \dfrac{2x - 1}{x-1} \)
- Answer
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- Yes.
- Yes.
- No.
- No.
Find the inverse of the function, if possible. Tell the domain and range of both functions.
- \( f(x) = x^2 + 1 \)
- \( f(x) = \sqrt{x} \)
- \( f(x) = \frac{x+2}{3} \)
- \( f(x) = \sqrt{2x-1} + 4 \)
- \( f(x) = \frac{1}{x} \)
- \( f(x) = |x| \)
- Answer
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- The domain of \(f\) is \( (-\infty,\infty) \) and the range is \( [1,\infty) \), but this function has no inverse.
- The domain and range of \(f\) are both \( [0,\infty) \). The inverse function is \( f^{-1}(x) = x^2,\) defined on domain \(x \geq 0 \).
- The domain and range of \(f\) are both \( (-\infty,\infty) \). The inverse function is \( f^{-1}(x) = 3x - 2 \), with same domain and range.
- The domain of \(f\) is \( x \geq \frac{1}{2} \) and the range is \( [4, \infty) \). The inverse function is \( f^{-1}(x) = \dfrac{ (x-4)^2 + 1}{2} \), defined only on \( x \geq 4\), with range \( \left[ \frac{1}{2}, \infty\right) \).
- The domain and range of \(f\) are both \( \{x \: | \: x \neq 0 \} \). The inverse function is \( f^{-1}(x) = \frac{1}{x} \), with same domain and range. Aka, this function is its own inverse!
- The domain of \(f\) is \( (-\infty, \infty) \) and the range is \( [0,\infty)\), but this function has no inverse.
1. The function \( k(m) = \frac{8}{5} m \) computes the approximate equivalent in kilometers per hour, given as input a speed in miles per hour. Find a function that converts (approximately) from kilometers per hour to miles per hour instead. Then convert \(80\) kph into mph.
2. The function \( F(C) = \frac{9}{5}C + 32 \) takes a Celsius temperature as input and computes the Fahrenheit equivalent as output. Find a function that converts from Fahrenheit to Celsius instead. Then convert \(95\) degrees Fahrenheit into Celsius.
3. The function \(V(h) = \pi r^2 h\) calculates the volume of a cylinder with (constant) radius \(r\), as a function of the height \(h\). Find a function that could take as input a desired volume and report as output the necessary height. Then, if a cylinder has radius \(2\) feet and a volume of \(16\pi\) cubic feet is desired, find the necessary height.
- Answer
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1. The function is \( m(k) = \frac{5}{8}k \), and \( m(80) = 50\) mph.
2. The function is \( C(F) = \frac{5}{9}(F - 32) \), and \(C(95) = 35\) degrees Celsius.
3. The function is \( h(V) = \frac{V}{\pi r^2} \), and \( h(16\pi) = \frac{16\pi}{\pi 2^2} = 4\) feet.