Section 6.13 Miscellaneous
These answers correspond to the problems in Section 3.8.
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Let \(f(x)=\ln x-cx^2\text{.}\) Note that the domain of \(f\) is the interval \((0,\infty )\) and that \(\ds \lim _{x\to \pm \infty }f(x)=-\infty\text{.}\) From \(\ds f'(x)=\frac{1}{x}-2cx\) it follows that there is a local (and absolute) maximum at \(\ds x=\frac{1}{\sqrt{2c}}\text{.}\) Since the function \(f\) is continuous on its domain, by the Intermediate Value Theorem, it will have a root if and only if \(\ds f\left( \frac{1}{\sqrt{2c}}\right) >0\text{.}\) Thus
\begin{equation*} \ln \frac{1}{\sqrt{2c}} -c\cdot\frac{1}{2c}>0\Leftrightarrow \ln \frac{1}{\sqrt{2c}}>\frac{1}{2}\Leftrightarrow c\lt \frac{1}{2e}\text{.} \end{equation*} Note that the function \(y=3x^3+2x+12\text{,}\) as a polynomial, is continuous on the set of real numbers. Also, \(\ds \lim _{x\to -\infty }(3x^3+2x+12)=-\infty\) and \(\ds \lim _{x\to \infty }(3x^3+2x+12)=\infty\text{.}\) By the Intermediate Value Theorem, the function has at least one root. Next, note that \(y'=6x^2+2>0\) for all \(x\in \mathbb{R}\text{.}\) This means that the function is increasing on its domain and therefore has only one root. (If there is another root, by Rolle's theorem there will be a critical number.)
Take \(y=x^3+9x+5\text{.}\)
\(a=-3\text{,}\) \(b=1\text{.}\) Solve \(f(10=6\) and \(f"(1)=0\text{.}\)
Note that the function \(f\) is continuous on its domain \(\mathbb{R}\backslash\{-1\}\text{.}\) Since \(\ds \lim _{x\to \infty }\frac{1}{(x+1)^2}=0\) and \(|\sin x|\leq 1\text{,}\) for all \(x\in \mathbb{R}\text{,}\) it follows that \(\ds \lim _{x\to \infty }f(x)=\lim _{x\to \infty }(-2x)=- \infty\text{.}\) Also, \(\ds f(0)=1>0\text{.}\) By the Intermediate Value Theorem, the function has at least one root in the interval \((0,\infty )\text{.}\) Next, note that \(\ds f'(x)=-\frac{2}{(x+1)^3} -2+\cos x\lt 0\) for all \(x\in (0,\infty)\text{.}\) This means that the function is decreasing on \((0,\infty)\) and therefore has only one root.
(a) From \(s'(t)=v(t)=96-2t\) it follows that the initial velocity is \(v(0)=96\) ft/sec. (b) The only critical number is \(t=48\text{.}\) By the second derivative test there is a local (and absolute) maximum there. (c) \(s(48)=2316\) feet.
(a) \(f'(3)=1\text{.}\) (b) \((fg)'(2)=f'(2)g(2)+f(2)g'(2)=-4\text{.}\) (c) \(f(0.98)\approx f(1)+f'(1)(0.98-1)=3+3(-0.02)=2.94\text{.}\) (d) There is a critical point at \((4,-1)\text{.}\)
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(a) The domain of \(f(x)=\arcsin x\) is the interval \([-1,1]\) and its range is \(\ds \left[ -\frac{\pi }{2},\frac{\pi }{2}\right]\text{.}\) (b) For \(x\in (-1,1)\) let \(y=\arcsin x\text{.}\) Then \(\sin y=x\) and \(y'\cos y=1\text{.}\) Since \(\ds y \in \left( -\frac{\pi }{2},\frac{\pi }{2}\right)\) it follows that \(\cos y>0\text{.}\) Thus \(\ds \cos y =\sqrt{1-\sin ^2y}=\sqrt{1-x^2}\text{.}\) (c) The domain of the function \(g\) is the set of all real numbers and its range is the set \(\ds \left[ -\frac{\pi }{2},\frac{\pi }{2}\right]\text{.}\) See Figure 6.13.1. (d) \(\ds g'(x)=\left\{ \begin{array}{rl} 1& \mbox{if } x\in (4n-1,4n+1), \ n\in \mathbb{Z}\\ -1& \mbox{if } x\in (4n+1,4n+3), \ n\in \mathbb{Z}\\ \end{array} \right.\text{.}\) The derivative of \(g\) is not defined if \(x=2n+1\text{,}\) \(n\in \mathbb{Z}\text{.}\) (e) The function \(F(x)=4x-2+\cos \left( \frac{\pi x}{2}\right)\) is continuous on the set of real numbers. From \(\ds \lim _{x\to \pm \infty }F(x)=\pm \infty\text{,}\) by the Intermediate Value Theorem, the function \(F\) has at least one root. From \(\ds F'(x)=1-\frac{2}{\pi }\sin \left( \frac{\pi x}{2}\right)>0\) we conclude that \(F\) is monotone.
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(a) \(\ds f'(x)=\frac{1-\ln x}{x^2}\text{.}\) (b) See Figure 6.13.2. (c) Since \(99>e\) we have that \(\ds \frac{\ln 99}{99}\lt \frac{\ln 101}{101}\text{.}\) This is the same as \(101\ln 99\lt 99\ln 101\text{.}\) Hence \(\displaystyle \ln 99^{101}\lt \ln 101^{99}\) and \(\displaystyle 99^{101}\lt 101^{99}\text{.}\)
(a) \(2x^3+x^2+5x+c, c\in \mathbb{R}\text{;}\) (b) \(\cosh (x)\text{;}\) (c) local minimum; (d) \(\displaystyle y^\prime =-x^4y^{-4}\text{;}\) (e) \((-3,0)\text{.}\)
(d) Take \(f(x)=x^3\) and the point \((0,0)\text{.}\) (e) For example, \(\ds f(x)=x-\frac{1}{x}\text{.}\)
(e) Take \(f(x)=-x^4\) and the point \((0,0)\text{.}\) (f) \(f(x)= x^2+3x+1\text{.}\)