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

The extreme f(x)=(x^2)/(1-x) is Minimum(0,0),Maximum(2,-4)

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

extreme points

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

Minimum (0, 0), Maximum (2, - 4)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Increasing

: 1 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

\frac{x ^{2}}{1 - x} : 0

Minimum (0, 0)

Plug

x = 2

into

\frac{x ^{2}}{1 - x} : - 4

Maximum (2, - 4)

Minimum (0, 0), Maximum (2, - 4)

f (x) = 4 - x^{2}, - 1 \leq x \leq 2

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

f (x) = \frac{3 x}{x ^{2} + 9}

f (x, y) = 10 x y - x^{3} - 5 y^{2}

f (x) = ln (7 - 2 x^{2})

What is the extreme f(x)=(x^2)/(1-x) ?
The extreme f(x)=(x^2)/(1-x) is Minimum(0,0),Maximum(2,-4)