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

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

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

extreme points

Minimum (0, 1)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 0

into

Minimum (0, 1)

f (x, y) = x^{2} + x y + y^{2} - 31 y + 320

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

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

f (x) = 6 x^{2} - 9 x + 5

f (x) = \frac{1}{x} [- 2.3]

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