What is the extreme f(x)= 3/5 (x^2-1)^{2/3} ?

The extreme f(x)= 3/5 (x^2-1)^{2/3} is Minimum(-1,0),Maximum(0, 3/5),Minimum(1,0)

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extreme f(x)= 3/5 (x^2-1)^{2/3}

extreme points

f (x) = \frac{3}{5} (x^{2} - 1)^{\frac{2}{3}}

Minimum (- 1, 0), Maximum (0, \frac{3}{5}), Minimum (1, 0)

Solution steps

f^{'} (x) = \frac{4 x}{5 ( 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

\frac{3}{5} (x^{2} - 1)^{\frac{2}{3}} : 0

Minimum (- 1, 0)

Plug

x = 0

into

\frac{3}{5} (x^{2} - 1)^{\frac{2}{3}} : \frac{3}{5}

Maximum (0, \frac{3}{5})

Plug

x = 1

into

\frac{3}{5} (x^{2} - 1)^{\frac{2}{3}} : 0

Minimum (1, 0)

Minimum (- 1, 0), Maximum (0, \frac{3}{5}), Minimum (1, 0)

f (x, y) = y^{2} + x y + 3 y + 2 x + 3

f (x) = \frac{5 e ^{4 x}}{5 x - 3}

f (x, y) = x y - x^{3} y^{2}

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

What is the extreme f(x)= 3/5 (x^2-1)^{2/3} ?
The extreme f(x)= 3/5 (x^2-1)^{2/3} is Minimum(-1,0),Maximum(0, 3/5),Minimum(1,0)