Q. 43 - Calculus

Question

Use the first derivative test to determine the local extrema of each function in Exercises 39- 50. Then verify your algebraic answers with graphs from a calculator or graphing utility.

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

Step-by-Step Solution

Answer

Ans: There are no critical points so no local extrema.

Step 1. Given Information:

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

Step 2. Finding the derivative of the function:

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

The derivative is defined and continuous everywhere except at $e^x=\frac{3}{2}$ , so the critical points of $f$ are just the points where $f\text{'}\left(x\right)=0$ ; that is.

$$\begin{gathered} \frac{-2e^x}{{\left(3-2e^x\right)}^2}=0 \\ {e}^x=0 \\ so there are no critical points. \\ \therefore there is no local extrema \end{gathered}$$

Step 3. Verifying algebraic answers with graphs :