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

The extreme f(x)=(2x^2)/(x-2) is Maximum(0,0),Minimum(4,16)

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

extreme points

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

Maximum (0, 0), Minimum (4, 16)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2,

Decreasing

: 2 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

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

Maximum (0, 0)

Plug

x = 4

into

\frac{2 x ^{2}}{x - 2} : 16

Minimum (4, 16)

Maximum (0, 0), Minimum (4, 16)

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

81 - x^{2}, - 9 \leq x \leq 9

\frac{( - 4 )}{x - 7}

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

g (x) = x 2 - x^{2}

What is the extreme f(x)=(2x^2)/(x-2) ?
The extreme f(x)=(2x^2)/(x-2) is Maximum(0,0),Minimum(4,16)