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

The extreme f(x)=(x^2)/(2-x) is Minimum(0,0),Maximum(4,-8)

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

extreme points

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

Minimum (0, 0), Maximum (4, - 8)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Increasing

: 2 < x < 4,

Decreasing

: 4 < x < \infty

Plug

x = 0

into

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

Minimum (0, 0)

Plug

x = 4

into

\frac{x ^{2}}{2 - x} : - 8

Maximum (4, - 8)

Minimum (0, 0), Maximum (4, - 8)

f (x) = \frac{2 x + 5}{3}, 0 \leq x \leq 5

y = - 5 (x - 4)^{4} + 2

x^{2} + 4 x y + y^{2} - 40 x - 56 y + 1

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

x y (1 - x - y)

What is the extreme f(x)=(x^2)/(2-x) ?
The extreme f(x)=(x^2)/(2-x) is Minimum(0,0),Maximum(4,-8)