What is the extreme f(x)= 1/(sqrt(1-x^2)) ?

The extreme f(x)= 1/(sqrt(1-x^2)) is Minimum(0,1)

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extreme f(x)= 1/(sqrt(1-x^2))

extreme points

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

Minimum (0, 1)

Solution steps

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

Find intervals:

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 1

Plug

x = 0

into

\frac{1}{1 - x ^{2}} : 1

Minimum (0, 1)

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

f (x) = \frac{- x - 8}{10 x ^{2} - 40 x + 27}

f (x, y) = 4 x^{2} + 9 y^{2} - 2 x - 18 x y^{2}

f (x) = \frac{x ^{3}}{3} - 10 x^{2} + 96 x + 100

f (x) = - x^{3} + 9 x^{2} + 120 x - 200, x \geq 5

What is the extreme f(x)= 1/(sqrt(1-x^2)) ?
The extreme f(x)= 1/(sqrt(1-x^2)) is Minimum(0,1)