Find all relative extrema of the function. у %3DIn(4 + 6x2- х3) a. y = -2 b. et 3.

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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How do I find the relative extrema of these equations?

Find all relative extrema of the function.
у %3DIn(4 + 6x2- х3)
a.
y =
-2
b.
et
3.
Transcribed Image Text:Find all relative extrema of the function. у %3DIn(4 + 6x2- х3) a. y = -2 b. et 3.
Expert Solution
Step 1

(a)

Consider the given function.

The given equation is y= In(4+6x² – x' ).
f(x) = In(4 + 6x –x')
-3x +12x
f'(x) =-
4+ 6x –
Find the critical points as follows,
f (x)= 0
-3x +12x
3
4 + 6x – x'
-3x? + 12x = 0
%3D
x = 0 or x = 4
So, the critical point is at x =4 and x = 0.
Step 2

Compute the second derivative of the function as follows.

f"(x)= d-3x? +12x
dx 4+6x-x
(-6x+ 12)(4 +6x -x')-(-3x +12x)(-3x +12x)
(4+ 6r* -x')*
-3x* + 24x – 72.x – 24x + 48
(4+6x -x')*
Substitute x = 4 and x = 0 in the obtained second derivative as follows.
At x = 4.
-3(4)' +24(4) - 72(4)° – 24(4)+48
(4+6(4) - (4')
f"(4) =
-432
362
3
<0
1(4)= la(4 +6(4)*-(4)')
= In 36
= 2 ln (6)
%3D
Step 3
Thus the function has relative maximum f(4)= 21n(6) that is (4,2ln(6)).
At x = 0.
r(0)=3(0)" + 24(0)' – 72(0)° – 24(0) + 48
(4+ 6(0)* - (0}')*
48
4?
= 3
S(0) = ln(4+ 6(0) - (0))
= In 4
%3D
= 2 In (2)
Thus the function has relative minimum f(0)=2 ln(2) that is (0,21n(2)).
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