Q. 34 - Calculus

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.)

$f\left(x\right)=e^x\left(x^2-x-1\right)$

Step-by-Step Solution

Answer

The function has a local minimum at $x=1$and has a local maximum at $x=-2.$

Step 1. Given Information.

The given function is $f\left(x\right)=e^x\left(x^2-x-1\right)$

Step 2. Critical points.

On differentiating the function, we get,

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

The critical points are points at which $f\text{'}\left(x\right)=0$

Therefore, critical points are $x=1\text{ and }x=-2$

Step 3. Second-Derivative Test.

Again differentiating the function, we get,

$$\begin{gathered} f^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}{e}^x\left(x^2-x-1\right)+e^x(2x-1) \\ =\left(x^2+3x-1\right)e^x \\ {f}^{\text{'}\text{'}}\left(1\right)=\left(1^2+3-1\right)e^1 \\ =3 e>0 \\ {f}^{\text{'}\text{'}}(-2)=\left((-2{)}^2+3(-2)-1\right)e^{-2} \\ =-3e^{-2}<0 \end{gathered}$$

Therefore, the function has a local maximum at $x=-2$ and local minimum at $x=1.$

Step 4. Verification.

The graph of the function is,

Which has local extrema at $x=-2,1.$