What is the extreme f(x)=\sqrt[3]{(x^2-1)^2} ?

The extreme f(x)=\sqrt[3]{(x^2-1)^2} is Minimum(-1,0),Maximum(0,1),Minimum(1,0)

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extreme f(x)=\sqrt[3]{(x^2-1)^2}

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

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

Minimum (- 1, 0)

Plug

x = 0

into

3 (x^{2} - 1)^{2} : 1

Maximum (0, 1)

Plug

x = 1

into

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

Minimum (1, 0)

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

f (x) = 8 - x

f (x) = 2 sin (3 x) - 5

(5 - x) (x + 1)^{2}

f (x) = 4 x^{7} + 9 x^{5} - 3 x^{4} + 2 x^{2} - 5

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

What is the extreme f(x)=\sqrt[3]{(x^2-1)^2} ?
The extreme f(x)=\sqrt[3]{(x^2-1)^2} is Minimum(-1,0),Maximum(0,1),Minimum(1,0)