(x, y) = (| relative maxima relative minima (х, у) %3D points of inflection (х, у) %- (smaller x-value) points of inflection (х, у) - (larger x-value)

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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A function and its first and second derivatives are given. Use these to find each of the following.
y =
x1/3(x - 8)
4(х — 2)
y'
3x2/3
4(x + 4)
y"
9x5/3
Find the relative maxima, relative minima, and points of inflection. (If an answer does not exist, enter DNE.)
relative maxima
(х, у) %3D
relative minima
(х, у) %3
points of inflection
(х, у) 3D
(smaller x-value)
points of inflection
(х, у) %3D
(larger x-value)
Transcribed Image Text:A function and its first and second derivatives are given. Use these to find each of the following. y = x1/3(x - 8) 4(х — 2) y' 3x2/3 4(x + 4) y" 9x5/3 Find the relative maxima, relative minima, and points of inflection. (If an answer does not exist, enter DNE.) relative maxima (х, у) %3D relative minima (х, у) %3 points of inflection (х, у) 3D (smaller x-value) points of inflection (х, у) %3D (larger x-value)
Expert Solution
Step 1

The given function is y=x1/3(x-8).

The first and the second derivative of y is y'=4(x-2)3x23                          ...... (1).

The second derivative of is y''=4(x+4)9x53                          ...... (2).

Compute the relative extrema as follows.

Substitute y'=0 in equation (1).

4(x-2)3x23=04(x-2)=0x=2

The y'=4(x-2)3x23is observed to be undefined at x=0.

Thus, the critical points are x=0 and x=2.

 

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