## 170- Quiz 8

Quiz 8 is here. Please remember that the second midterm is this Wednesday.

Solutions follow.

Problem 1 asks us to determine the local maxima and minma of $f(x)=e^x/x$.

We first note that the function $f$ is continuous and differentiable everywhere, except at $x=0$, where $f$ is not even defined. We have $\displaystyle f'(x)=\frac{e^x\cdot x-e^x\cdot1}{x^2}=\frac{e^x(x-1)}{x^2},$

using the quotient rule. Since $e^x>0$ for all $x$, it follows that the only solution to $f'(x)=0$ is $x=1$. Since $f'(x)<0$ is $x<1$ and $f'(x)>0$ is $x>1$, it follows that the only extreme point of $f$ occurs at $(1,f(1))=(1,e)$, and it is a minimum.

This completes the problem, but it may be instructive to analyze the function a bit more.

Note that $\displaystyle \lim_{x\to0^+}\frac{e^x}x=+\infty$ and $\displaystyle\lim_{x\to0^-}\frac{e^x}x=-\infty$,

so the $y$-axis is a vertical asymptote at $x=0$.

Also, $\displaystyle\lim_{x\to-\infty}\frac{e^x}x=0$

so the $x$-axis is a horizontal asymptote as $x\to-\infty$, while, using L’Hôpital’s rule, we have $\displaystyle \lim_{x\to+\infty}\frac{e^x}x=\lim_{x\to+\infty}e^x=+\infty$.

Note that $f(x)<0$ if $x<0$ and $f(x)>0$ if $x>0$. Also, from the sign of the derivative, $f$ is decreasing in $(-\infty,0)$ and in $(0,1)$, and it is increasing in $(1,\infty)$.

Finally, $\displaystyle f''(x)=\left(\frac{e^x(x-1)}{x^2}\right)'=\frac{(e^x(x-1))'x^2-e^x(x-1)2x}{x^4}$ $\displaystyle =\frac{(e^x(x-1)+e^x\cdot1)x^2-e^x(2x^2-2x)}{x^4}$ $\displaystyle =\frac{e^x(x^3-x^2+x^2-2x^2+2x)}{x^4}=\frac{e^x(x^3-2x^2+2x)}{x^4}$ $\displaystyle =\frac{e^x(x^2-2x+2)}{x^3}=\frac{e^x((x-1)^2+1)}{x^3}.$

Clearly, $f''(x)>0$ if $x>0$ and $f''(x)<0$ if $x<0$.

This means that $f$ is concave down in $(-\infty,0)$ and concave up in $(0,\infty)$.

Combining these observations allows us to sketch $f$ with reasonable accuracy. The graph of the function is shown below. (Click the graph to enlarge.) $*\ *\ *$

Problem 2 asks us to consider the function $f(x)=-x^4+20x^2-64$, and to indicate the intervals where it is increasing or decreasing, concave up, or concave down.

This function and its derivative are polynomials, so they are defined for all values of $x$. First, we identify the values $x$ where $f'(x)=0$.

We have $f'(x)=-4x^3+40x=-4x(x^2-10)$ $=-4x(x-\sqrt{10})(x+\sqrt{10}).$

Next, we identify the values $x$ where $f''(x)=0$.

We have $f''(x)=(-4x^3+40x)'=-12x^2+40,$

so $f''(x)=0\Longleftrightarrow 40=12x^2$, or $\displaystyle \frac{40}{12}=\frac{10}3=x^2$, so $x=\sqrt{10/3}$ or $x=-\sqrt{10/3}$.

Note that $-\sqrt{10}<-\sqrt{10/3}<0<\sqrt{10/3}<\sqrt {10}$, i.e., the zeroes of $f'$ and of $f''$ intertwine. (This is actually a general phenomenon.)

To determine the concavity of $f$ in the intervals determined by the zeroes of $f''$, we evaluate $f''$ at points in the intervals $(-\infty,-\sqrt{10/3}),$ $(-\sqrt{10/3},\sqrt{10/3}),$ and $(\sqrt{10/3},+\infty)$. Thanks to the second derivative test, we can save some time, and use these evaluations to (simultaneously) determine the extrema of $f$ and the intervals where $f$ increases and decreases.

We have $f''(-\sqrt{10})=-120+40<0$, so $(-\sqrt{10},f(-\sqrt{10}))=(-\sqrt{10},36)$ is a local maximum of $f$, and $f$ is concave down in $(-\infty,-\sqrt{10/3})$.

Similarly, $f''(\sqrt{10})<0$ so $(\sqrt{10},f(\sqrt{10}))=(\sqrt{10},36)$ is a local maximum of $f$, and $f$ is concave down in $(\sqrt{10/3},\infty)$.

Finally, $f''(0)=40>0$, so $(0,f(0))=(0,-64)$ is a local minimum of $f$ and $f$ is concave up in $(-\sqrt{10},\sqrt{10})$.

Moreover (due to the placement of its extreme points) we can also deduce that $f$ is increasing in $(-\infty,-\sqrt{10}),$ decreasing in $(-\sqrt{10},0),$ increasing in $(0,\sqrt{10}),$ and decreasing in $(\sqrt{10},\infty)$, so $x=\pm\sqrt{10}$ give us not just local maxima but global maxima of $f.$

The graph of $f$ is shown below. (Click the image to enlarge).