Question

Use the second-derivative test to determine the local extrema of each function $f$in Exercises $29-40$. If the second-derivative test fails, you may use the first-derivative test. Then verify your algebraic answers with graphs from a calculator or graphing utility. (Note: These are the same functions that you examined with the first-derivative test in Exercises 39–50 of Section 3.2.)

$f\left(x\right)=\frac{1}{3-2e^x}$

Short answer

There is no local extrema.

Step-by-step solution

Step 1. Given Information.

The given function is$f\left(x\right)=\frac{1}{3-2e^x}.$

Step 2. Critical points.

On differentiating the function, we get,

$$\begin{gathered} f\left(x\right)=\frac{1}{3-2e^x} \\ {f}^{\text{'}}\left(x\right)=\frac{d}{dx}\frac{1}{3-2e^x} \\ =\frac{\left(\frac{d}{dx}1\right)\left(3-2e^x\right)-1\left(\frac{d}{dx}\left(3-2e^x\right)\right)}{{\left(3-2e^x\right)}^2} \\ =\frac{-\left(\frac{d}{dx}3-2\frac{d}{dx}{e}^x\right)}{{\left(3-2e^x\right)}^2} \\ =\frac{2e^x}{{\left(3-2e^x\right)}^2} \end{gathered}$$

Now, there is no value of $x$for which $f\text{'}\left(x\right)$is $0.$

Therefore, there is no critical point.

Therefore, there is no local extrema.

Step 3. Verification.

The graph of the function is

It has no local extrema.