## 170- Quiz 7

Quiz 7 is here.

Solutions follow.

Problem 1 asks for the derivative of $\arctan(e^x)$.

Recall that $\displaystyle (\arctan x)'=\frac1{1+x^2}$.

If you do not remember this, it can be found easily: If $y=\arctan x$, then $\tan y=x$, so $(\sec^2 y)y'=1$ or $\displaystyle y'=\frac1{\sec^2y}$. But $\sin^2y+\cos^2y=1$, so (dividing this identity by $\cos^2y$) $\tan^2y+1=\sec^2y$. Now, since $\tan y=x$, we get $\tan^2y=x^2$ and $\sec^2y=1+x^2$. Finally, $\displaystyle y'=\frac1{1+x^2}$.

Using the chain rule, we obtain $\displaystyle (\arctan(e^x))'=\frac{e^x}{1+(e^x)^2}=\frac{e^x}{1+e^{2x}}$.

Problem 2 defines $f(\theta)=\cos^2(\theta)-2\sin(\theta)$, and asks for the intervals where $f$ is increasing and where $f$ is decreasing within ${}[0,2\pi]$. Then, it asks to use this information to find the local maxima and minima of $f$.

We know that is $f$ is differentiable, then $f$ increases where $f'$ is positive, and decreases where $f$ is negative. Since our function $f(\theta)$ is certainly differentiable, we may start by finding the points where $f'$ is 0: $f'(\theta)=2\cos(\theta)(-\sin(\theta))-2\cos(\theta)=-2\cos(\theta)(\sin(\theta)+1)$, so $f'(\theta)=0$ iff $\cos(\theta)=0$ or $\sin(\theta)+1=0$. In the first case, $\displaystyle\theta=\frac{\pi}2$ or $\displaystyle \frac{3\pi}2$ (remember that we only consider values of $\theta$ in ${}[0,2\pi]$). In the second case, $\sin(\theta)=-1$ or $\displaystyle \theta=-\frac{3\pi}2$. Below I indicate the distinguished points within the interval ${}[0,2\pi]$: We have identified 3 intervals in ${}[0,2\pi]$: The interval where $\theta<\pi/2$, the one with $\pi/2<\theta<3\pi/2$, and the one where $3\pi/2<\theta$. In each of these intervals, the sign of $f'(\theta)$ does not change, so to determine it, it suffices to evaluate $f'(\theta_0)$ for specific points $\theta_0$ chosen in each interval. In the first one, we can actually take $\theta_0=0$ because 0 is not a critical point of $f$ even though it is a distinguished point of our analysis, being an endpoint. We have $f'(0)=-2\cdot1\cdot(0+1)=-2<0$.

It follows that $f$ is decreasing in $(0,\pi/2)$.

Since $f(2\pi)=f(0)$, we can take $\theta_0=2\pi$ in the last interval, and it follows that $f$ is decreasing in $(3\pi/2,2\pi)$.

Finally, we can take $\theta_0=\pi$ in the middle interval, and we see that $f'(\pi)=-2(-1)(0=1)=2>0$, and therefore $f$ is increasing in $(\pi/2,3\pi/2)$.

We have $f(0)=1$, $f(\pi/2)=-2$, $f(3\pi/2)=2$ and $f(2\pi)=1.$ From our analysis, we conclude that:

• $(0,1)$ is a local maximum.
• $(\pi/2,-2)$ is a local minimum. (Actually, this is the global minimum of $f$.)
• $(3\pi/2,2)$ is a local maximum. (Actually, this is the global maximum of $f$.)
• $(2\pi,1)$ is a local minimum.

The graph of $f$, shown below, confirms our findings. Click the image above to enlarge.