Exam 3 solutions
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$f(x)=\frac{x^2-1}{x^3}$

**(a) State the domain of the function**

All reals except $x=0$

**(b) State all the x- and y-intercepts.**

(-1,0), (1,0)

**(c\)State all the equations of horizontal asymptotes**

$y=0$

**(d) State all equations of vertical asymptotes.**

$x=0$

**(e) List the interval(s) of increase.**

$(-\sqrt{3},0)$, $(0,\sqrt{3})$

**(f) State the $x$-values of all local maxima.**

$$x=\sqrt{3}$$

**(g) State the $x$-values of all local maxima.**

$$x=-\sqrt{3}$$

**(h) State the intervals on which the graph is concave down.**

$(-\infty,-\sqrt{6})$, $(0,\sqrt{6})$

**(f) State the $x$-values of all inflection points.**

$$x=\pm\sqrt{6}$$

(j)

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**2. Find the global minimum and global maximum value of the function `g\left(x\right)=x-2\cos\left(x\right)` on the interval `\left[0,2\pi\right]`. Show all your work and explain your full thought process. You have the option to draw using the sketch tool on the left. Be sure to end with a sentence summarizing your answer.**
$g\left(x\right)=x-2\cos\left(x\right)$
$g'(x)=1+2\sin(x)$. $g'$ is defined everywhere. Set $g'=0$ and solve. $-1/2=\sin(x)$, so $x=7\pi/6,11\pi/6$.

Compare
$g(0)=-2$
$g(2\pi)=2\pi-2$
$g(7\pi/6)=7\pi/6+\sqrt{3}$
$g(11\pi/6)=11\pi/6-\sqrt{3}$

On the interval $[0,2\pi]$, $g$ has a global max value of $g(7\pi/6)=7\pi/6+\sqrt{3}$ and a global min value of $g(0)=-2$.

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**3. You have been tasked with designing a rectangular pavilion that will be subdivided into three identical rectangular rooms with all walls made of curtains as pictured above. The total area must be 72 square meters.**

**What should the outer dimensions of the pavilion be to minimize the total length of the curtained walls?**

Let $x$ be the horizontal edge and $y$ be the vertical edge. Then $xy=72$, so $y=72/x$.

We want to minimize the perimeter $P=2x+4y=2x+4(72x^-1)$.
$P'=2-288x^{-2}$. Note that $P'$ is defined for all $x>0$. Set $P=0$ and solve. $2=\frac{288}{x^{-2}}$ so $x=\pm 12$. Only $x=12$ is in the interval $(0,\infty)$.

*1st derivative test*
$P'(1)<0$ and $P'(20)>0$. Thus $P$ goes from decreasing to increasing at $x=12$. Then $P$ is minimized when $x=12$.

The outer dimensions should be 12m x 6m.

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**4.Town A is 30 km North of Town B. A biker rides East from Town A at a speed of 12 km/h at noon, and a biker rides West from Town B at a speed of 8 km/h at noon. At 2:00, how quickly is the distance between the bikers increasing?**

Let $z$ be the distance between the riders, and let $x$ be the horizontal distance. (Some students let $x=a+b$ and found $\frac{da}{dt}$ and $\frac{db}{dt}$) Then
* $x^2+30^2=z^2$.
* We know $\frac{dx}{dt}=20$km/h.
* We want $\frac{dz}{dt}$ at 2:00 (when $x=40$km and $z=50$km).

$2x\frac{dx}{dt}=2z\frac{dz}{dt}$

At 2:00 we get $2(40)(20)=2(50)\frac{dz}{dt}$ so $\frac{dz}{dt}=16$km/h

At 2:00, the distance between the bikers is increasing at a rate of 16km/h.