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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

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

Minimum (- 1, 0)

Plug

x = 0

into

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

Maximum (0, 1)

Plug

x = 1

into

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

Minimum (1, 0)

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

f (x) = \frac{1}{4 x ^{2} - 17 x + 4}

\frac{( 2 x - 1 )}{2 x + 5}

f (x) = - x^{4}

f (x) = \frac{( x - 9 )}{( x + 9 )}

2 x - y = 5

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