Answered: Let f(x) = 3x^4 − 2x^3. Find all the…

College Algebra

1st Edition

ISBN:9781938168383

Author:Jay Abramson

Publisher:Jay Abramson

Chapter3: Functions

Section3.3: Rates Of Change And Behavior Of Graphs

Problem 3SE: How are the absolute maximum and minimum similar to and different from the local extrema?

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Question

Let f(x) = 3x^4 − 2x^3. Find all the critical points and determine if they are relative maximums, relative minimums, or neither. Then find the absolute max and absolute min on the interval [−1, 1]. Show and explain all your work.

Expert Solution

Step 1

First find the critical point.

Obtain the first derivative of the given function as follows.

$f' (x) = 12 x^{3} - 6 x^{2}$

Equate the first derivative to zero and solve for x.

$12 x^{3} - 6 x^{2} = 0 x^{2} (12 x - 6) = 0 x^{2} = 0 o r 12 x - 6 = 0 x = 0 o r 12 x = 6 x = 0 o r x = \frac{1}{2}$

Therefore, the critical points are $x = 0 o r x = \frac{1}{2}$ .

These points divide the interval [-1,1] into three intervals $[- 1, 0], [0, \frac{1}{2}] a n d [\frac{1}{2}, 1]$ .

Choose test points from each interval and substitute those points in the first derivative to find the sign.

Interval	$[- 1, 0]$	$[0, \frac{1}{2}]$	$[\frac{1}{2}, 1]$
Test point	$- \frac{1}{2}$	$\frac{1}{4}$	$\frac{3}{4}$
Value of the first derivative	-3	$- \frac{3}{16}$	$\frac{27}{16}$
Sign	Negative	Negative	Positive

Note that, at x=0 there is no sign change and at x= $\frac{1}{2}$ the sign of the derivative changes from negative to positive.

Therefore, there is a relative minimum at x= $\frac{1}{2}$ and a saddle point at x=0.

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