Question
Asked Oct 20, 2019
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Let f (x) = x1/3(4x + 6)2/3.   Find f ' (x) and the critical numbers for f.

f ' (x) = 
Ax + B
3 xC (4x + 6)D
, where
A =
B =
C =
D =

The critical numbers of f are (list in increasing order):
x =
x =
x =


(Enter DNE in any unused answer blank.)

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Expert Answer

Step 1

Use product rule to find f'(x). 

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2 f(x)=x3 (4x+6) 2 2 f'(x)=(x3)(4x+6)3 x (4x+6) 3 2 2 1 2 (4x+6)x 3 f'(x)=X 3 (4x+6) 2 (4x+6) X f'(x)= 2 3 3x 3(4x+6)

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Step 2

Simplify it using LCD

Answer: A=5, B=6, ...

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2 3 X (4x+6) f'(x)= 2 Зx 3(4x+6) 2 (4х+6) ' (4х+6)* +x *x 3 f'(x)= 2 Зх 3 (4x+6) 4x+6+х г')- 2 Зх (4х+6) 3 5х+6 f'(x) 2 Зх 3 (4х+6) - |.

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