Q.49 - Calculus

Question

Use a sign chart for $f\text{'}\text{'}$ to determine the intervals on which each function f in Exercises 41–52 is concave up or concave down, and identify the locations of any inflection points. Then verify your algebraic answers with graphs from a calculator or graphing utility 

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

Step-by-Step Solution

Answer

Inflection point is $x=\frac{\sqrt{2}+2}{2},-\frac{\sqrt{2}-2}{2}$, concave up on $\left(-\infty ,-\frac{\sqrt{2}-2}{2}\right)\left(\frac{\sqrt{2}+2}{2},\infty \right)$ and concave down on $\left(-\frac{\sqrt{2}-2}{2},\frac{\sqrt{2}+2}{2}\right).$

Step 1. Given Information.

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

Step 2. Second-derivative.

On differentiating,

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d}{dx}{e}^{1+2x-x^2} \\ =e^{1+2x-x^2}(2-2x) \\ {f}^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}{e}^{1+2x-x^2}(2-2x) \\ =2\left(2x^2-4x+1\right)e^{1+2x-x^2} \end{gathered}$$

Step 3. Sign chart.

Now, $f\text{'}\text{'}\left(x\right)=0 at x=\frac{\sqrt{2}+2}{2},-\frac{\sqrt{2-2}}{2}.$

Therefore, these are the inflection points.

The sign chart will be:

Concave up on $\left(-\infty ,-\frac{\sqrt{2}-2}{2}\right)\left(\frac{\sqrt{2}+2}{2},\infty \right)$

Concave down on $\left(-\frac{\sqrt{2}-2}{2},\frac{\sqrt{2}+2}{2}\right)$

Step 4. Verification.

The graph of the function is,