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

The extreme f(x)=(2x^2)/(x-4) is Maximum(0,0),Minimum(8,32)

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

extreme points

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

Maximum (0, 0), Minimum (8, 32)

Solution steps

f^{'} (x) = \frac{2 ( x ^{2} - 8 x )}{( x - 4 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 4,

Decreasing

: 4 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = 0

into

\frac{2 x ^{2}}{x - 4} : 0

Maximum (0, 0)

Plug

x = 8

into

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

Minimum (8, 32)

Maximum (0, 0), Minimum (8, 32)

What is the extreme f(x)=(2x^2)/(x-4) ?
The extreme f(x)=(2x^2)/(x-4) is Maximum(0,0),Minimum(8,32)