Miscellaneous

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Section 6.13 Miscellaneous

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These answers correspond to the problems in Section 3.8.

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 $(4,-1)\text{.}$
(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.

Figure $f(x)=\arcsin(\sin x)$
(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 $\displaystyle 99^{101}\lt 101^{99}\text{.}$

Figure $\ds f(x)=\frac{\ln x}{x}$
(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) $f(x)= x^2+3x+1\text{.}$