Q. 86 - Calculus

Question

In Exercises 83–86, use the given derivative f' to find any local extrema and inflection points of f and sketch a possible graph without first finding an formula for f . 

$\mathrm{f}\text{'}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}\left(\mathrm{x}+4\right)$

Step-by-Step Solution

Answer

A critical point is $\mathrm{x} =- 4$

Step 1: Given information

The function's name is $\mathrm{f}\text{'}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}\left(\mathrm{x}+4\right)$

Step 2: Calculations

Consider the following factors while determining the important points:

$\begin{array}{c}\mathrm{f}\text{'}\left(\mathrm{x}\right)=0\\ {\mathrm{e}}^{\mathrm{x}}\left(\mathrm{x}+4\right)=0\\ \begin{array}{rcl}\left(\mathrm{x}+4\right)& =& 0\\ \mathrm{x}& =& 4\end{array}\end{array}$

As a result,  A critical point is $\mathrm{x}=-4$

Then take a look at the table below.

Interval	$(-\mathrm{\infty },-4)$	$(-4,\mathrm{\infty })$
k	-5	0
Value of the test for $\mathrm{f}\text{'}\left(\mathrm{k}\right)$	-	+
sign of $\mathrm{f}\text{'}\left(\mathrm{x}\right)$	${\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)<0$	${\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)>0$
Conclusion	Decreasing	Increasing

Therefore,  The crucial point is$\mathrm{x}=-4$.

Then there's the graph of f,