M119 In-class Review Answers

$\hsize=6truein \newcount\qno \def\qn {\par\advance\qno by1\noindent\strut \hbox to\parindent{\the\qno.\hfil}\hangindent\parindent\ignorespaces} \font\bg=cmbx10 scaled \magstep2 \font\hg=cmr7 scaled 2985 \vglue 0pt \rm \qn $\displaystyle {3\over 2\sqrt x}-{2\over x^3}$$
$\qn $\displaystyle {2(x^2+9)-(2x+5)(2x)\over (x^2+9)^2}$$
$\qn %$h'(x)=5(6x^3-27)^4(18x^2)$ and $h''(x)=20(6x^3-27)^3(18x^2)^2+5(6x^3-27)^4(36x)$$
$\qn $\displaystyle f'(x)=\lim_{h\to 0}{f(x+h)-f(x)\over h}$$
$\qn $29/7$$
$\qn $1/4$$
$\qn $-5$$
$\qn $\displaystyle x-1={1\over 2}(y-1)$$
$\qn $f'(x)=4x^3-12x^2=4x^2(x-3)$ so increasing when $x>3$ and decreasing when $x<3$. $f''(x)=12x^2-24x=12x(x-2)$ so concave up when $x>2$ or $x<0$ and concave down when $0<x<2$.$
$\qn $f(x)$ is discontinuous at $x=-1$ and $x=1$.$
$\qn $\displaystyle f'(x)={2(x^2+4)-4x^2\over (x^2+4)^2}= {-2(x-2)(x+2)\over (x^2+4)^2}=0$ implies $x=2$ or $-2$. Check the endpoints $-2$ and $3$ as well. $f(2)=1/2$, $f(-2)=-1/2$, $f(3)=6/13$. Hence the maximum is $1/2$ and the minimum is $-1/2$.$
$\qn$
A graph of f'(x) is here.

$\qn $P(x)=xp-C(x)=x(200-3x)-75-80x+x^2=-2x^2+120x-75$ so that $P'(x)=-4x+120=0$ implies $x=30$. Check the endpoints as well at $0$ and $40$. $P(0)=-75$, $P(30)=1725$, $P(40)=1525$. Hence the maximum profit occurs at $x=30$ with a price of $p=110$.$
$\qn $\displaystyle\lim_{h\to 0} {f(4+h)-f(4)\over h}= f'(4)={1\over 2\sqrt 4}=1/4$.$
$\qn $(f\circ g)'(7)=f'\big(g(7)\big)g'(7) =f'(3)4=8$$