Q. 7 - Calculus

Question

Sign analyses for second derivatives: Repeat the instructions of the previous block of problems, except find sign intervals for the second derivative $f\text{'}\text{'}$ instead of the first derivative.

$f\left(x\right)=\frac{\sin x}{e^x}$

Step-by-Step Solution

Answer

The second derivative of the given function is the local maximum at $x=\frac{\mathrm{\pi }}{4}.$

Step 1. Given Information.

The given function is $f\left(x\right)=\frac{\sin x}{e^x}.$

Step 2. Find the second derivative.

To find the second derivative, first, we find the first derivative then the second.

So,

$$\begin{gathered} f\left(x\right)=\frac{\sin x}{e^x} \\ f\text{'}\left(x\right)=\frac{e^x\left(\cos x\right)-\sin x\left(e^x\right)}{{\left(e^x\right)}^2} \\ f\text{'}\left(x\right)=\frac{\cos x-\sin x}{e^x} \\ f\text{'}\text{'}\left(x\right)=\frac{e^x(-\sin x-\cos x)-e^x(\cos x-\sin x)}{e^{2x}} \\ f\text{'}\text{'}\left(x\right)=\frac{-2e^x\cos x}{e^{2x}} \\ f\text{'}\text{'}\left(x\right)=\frac{-2\cos x}{e^x} \end{gathered}$$

Step 3. Find sign intervals for the second derivative.

To find the sign intervals let's find the critical points from the first derivative.

So,

$$\begin{gathered} f\text{'}\left(x\right)=\frac{e^x\left(\cos x\right)-\sin x\left(e^x\right)}{\left(e^{2x}\right)} \\ 0=\frac{e^x\left(\cos x\right)-\sin x\left(e^x\right)}{\left(e^{2x}\right)} \\ 0=e^x\left(\cos x-\sin x\right) \\ 0=\cos x-\sin x \\ 1=\tan x \\ x=\frac{\mathrm{\pi }}{4} \end{gathered}$$

Thus, the critical point is $x=\frac{\mathrm{\pi }}{4}.$

Now, at $x=\frac{\mathrm{\pi }}{4}, f\text{'}\text{'}\left(x\right)<0.$