Chapter 10: Q. 86 (page 777)

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 .
$f' (x) = e^{x} (x + 4)$

Short Answer

Expert verified

A critical point is $x = - 4$

Step by step solution

Given information

The function's name is $f' (x) = e^{x} (x + 4)$

Calculations

Consider the following factors while determining the important points:

$\begin{matrix} f' (x) = 0 \\ e^{x} (x + 4) = 0 \\ \begin{array}{rcl} (x + 4) & = & 0 \\ x & = & 4 \end{array} \end{matrix}$

As a result, A critical point is $x = - 4$

Then take a look at the table below.

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

Therefore, The crucial point is $x = - 4$ .

Then there's the graph of f,

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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 .
$f' (x) = e^{x} (x + 4)$

Short Answer

Step by step solution

Given information

Calculations

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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 .f'x=exx+4

Short Answer

Step by step solution

Given information

Calculations

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Mechanics Maths

Calculus

Pure Maths

Theoretical and Mathematical Physics

Geometry

Study anywhere. Anytime. Across all devices.

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 .
$f' (x) = e^{x} (x + 4)$