What is the extreme f(x)=((4x^2))/(x-2) ?

The extreme f(x)=((4x^2))/(x-2) is Maximum(0,0),Minimum(4,32)

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extreme f(x)=((4x^2))/(x-2)

extreme points

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

Maximum (0, 0), Minimum (4, 32)

Solution steps

f^{'} (x) = \frac{4 ( 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{4 x ^{2}}{x - 2} : 0

Maximum (0, 0)

Plug

x = 4

into

\frac{4 x ^{2}}{x - 2} : 32

Minimum (4, 32)

Maximum (0, 0), Minimum (4, 32)

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

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

x^{\frac{1}{9}} + 1

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

f (x, y) = x y + \frac{8}{x} + \frac{8}{y}

What is the extreme f(x)=((4x^2))/(x-2) ?
The extreme f(x)=((4x^2))/(x-2) is Maximum(0,0),Minimum(4,32)