f(x) = x³/5(7 - x) Solution First note that the domain of f is R. The Product Rule gives the following. f'(x) = x³/5-1 (³) -5V-5 5√² (-8xVx³ +21 √√x³) 5.x Therefore f'(x) = 0 if 21 - -1 y [The same result could be obtained by first writing f(x) = 7x3/5 - x8/5.] 8 7 6 5 4 3 2 1 3 ) + (7-X 5√² 1 2 +3(7x) 5.1.² - x)) 3 3(7x) 4 5 = 0, that is, x = 6 7 8 and f'(x) does not exist when x = Thus the critical numbers are (larger) and (smaller).
f(x) = x³/5(7 - x) Solution First note that the domain of f is R. The Product Rule gives the following. f'(x) = x³/5-1 (³) -5V-5 5√² (-8xVx³ +21 √√x³) 5.x Therefore f'(x) = 0 if 21 - -1 y [The same result could be obtained by first writing f(x) = 7x3/5 - x8/5.] 8 7 6 5 4 3 2 1 3 ) + (7-X 5√² 1 2 +3(7x) 5.1.² - x)) 3 3(7x) 4 5 = 0, that is, x = 6 7 8 and f'(x) does not exist when x = Thus the critical numbers are (larger) and (smaller).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![Solution
First note that the domain of f is R. The Product Rule gives the following.
f'(x)
f(x) = x³/5(7 − x)
=
x3/5 −1
=
- (x (²)
-5√x-5
(-8xVx³ +21√x³
5√1²
۱
5x
Therefore f'(x) = 0 if 21 -
-1
[The same result could be obtained by first writing f(x) 7x3/5 - x8/5.]
y
8
7
6
5
4
3
2
1
3
) + (7 - x0)( 5√²
3(7 - x)
)+(1DV=
+ 3(7 - x))
1
2
3
4
5
= 0, that is, x =
6
8
and f'(x) does not exist when x =
Thus the critical numbers are
(larger) and
(smaller).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9b15a2e4-841f-4430-ad46-9439390435a9%2F2df08761-81d0-4778-bbd9-b72c6a5e3bb4%2F1ri0yn8_processed.png&w=3840&q=75)
Transcribed Image Text:Solution
First note that the domain of f is R. The Product Rule gives the following.
f'(x)
f(x) = x³/5(7 − x)
=
x3/5 −1
=
- (x (²)
-5√x-5
(-8xVx³ +21√x³
5√1²
۱
5x
Therefore f'(x) = 0 if 21 -
-1
[The same result could be obtained by first writing f(x) 7x3/5 - x8/5.]
y
8
7
6
5
4
3
2
1
3
) + (7 - x0)( 5√²
3(7 - x)
)+(1DV=
+ 3(7 - x))
1
2
3
4
5
= 0, that is, x =
6
8
and f'(x) does not exist when x =
Thus the critical numbers are
(larger) and
(smaller).
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