What is the extreme f(x)=(x+1)^{2/3}(x-3)^2 ?

The extreme f(x)=(x+1)^{2/3}(x-3)^2 is Minimum(-1,0),Maximum(0,9),Minimum(3,0)

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extreme f(x)=(x+1)^{2/3}(x-3)^2

extreme points

f (x) = (x + 1)^{\frac{2}{3}} (x - 3)^{2}

Minimum (- 1, 0), Maximum (0, 9), Minimum (3, 0)

Solution steps

f^{'} (x) = \frac{8 x ( - 3 + x )}{3 ( 1 + x ) ^{\frac{1}{3}}}

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 1

into

(x + 1)^{\frac{2}{3}} (x - 3)^{2} : 0

Minimum (- 1, 0)

Plug

x = 0

into

(x + 1)^{\frac{2}{3}} (x - 3)^{2} : 9

Maximum (0, 9)

Plug

x = 3

into

(x + 1)^{\frac{2}{3}} (x - 3)^{2} : 0

Minimum (3, 0)

Minimum (- 1, 0), Maximum (0, 9), Minimum (3, 0)

f (x) = \frac{1}{4} x^{4} + \frac{4}{3} x^{3} + 2 x^{2}

f (x) = (1 - x^{2})^{2}

f (x, y) = 2 x y + x^{2} y - y

f (x) = 5 x^{\frac{2}{3}} - 2 x^{\frac{5}{3}}

f (x) = ln (x^{2} + x + 1)

What is the extreme f(x)=(x+1)^{2/3}(x-3)^2 ?
The extreme f(x)=(x+1)^{2/3}(x-3)^2 is Minimum(-1,0),Maximum(0,9),Minimum(3,0)