Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Question

Chapter 5.1, Problem 28E

To determine

To find: theextreme values of the function and where they occur.

Expert Solution & Answer

Answer to Problem 28E

the minimum value is −7.22 at x=−2+10 and maximum value is 10 at x=0 .

Explanation of Solution

Given information:

The function is y=32x4+4x3−9x2+10

Calculation: the function is defined for each value of x,

The function domain is R , so the domain of function has no end points, so the extreme points occur at critical points,

To find the extreme values of function, first find its derivative,

y=32x4+4x3−9x2+10y'=3x3+12x2−18x [differentiate and simplify]

For critical points equal the derivative of function to zero,

3x3+12x2−18x=0x=0,−2±10 [simplify]

Now substitute in function,

y=32x4+4x3−9x2+10y=10 [substitute x=0 ]y=5.36 [substitute x=−2+10]y=−7.22 [substitute x=−2+10]

Thus, the minimum value is −7.22 at x=−2+10 and maximum value is 10 at x=0 .

Chapter 5.1, Problem 27E

Chapter 5.1, Problem 29E

Chapter 5 Solutions

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

Ch. 5.1 - Prob. 1QR Ch. 5.1 - Prob. 2QR Ch. 5.1 - Prob. 3QR Ch. 5.1 - Prob. 4QR Ch. 5.1 - Prob. 5QR Ch. 5.1 - Prob. 6QR Ch. 5.1 - Prob. 7QR Ch. 5.1 - Prob. 8QR Ch. 5.1 - Prob. 9QR Ch. 5.1 - Prob. 10QR

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