What is the extreme f(x)=(2x^2)/(x-8) ?

The extreme f(x)=(2x^2)/(x-8) is Maximum(0,0),Minimum(16,64)

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extreme f(x)=(2x^2)/(x-8)

extreme points

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

Maximum (0, 0), Minimum (16, 64)

Solution steps

f^{'} (x) = \frac{2 ( x ^{2} - 16 x )}{( x - 8 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 8,

Decreasing

: 8 < x < 16,

Increasing

: 16 < x < \infty

Plug

x = 0

into

\frac{2 x ^{2}}{x - 8} : 0

Maximum (0, 0)

Plug

x = 16

into

\frac{2 x ^{2}}{x - 8} : 64

Minimum (16, 64)

Maximum (0, 0), Minimum (16, 64)

\frac{x ^{2}}{x ^{2} + 4}

8 x y

f (x, y) = e^{x^{2} + y^{2} - 4 x}

f (x) = 6 x^{4} - 32 x^{3} + 48 x^{2} - 7

f (x) = (3 t^{2}), (0 \leq t \leq 15)

What is the extreme f(x)=(2x^2)/(x-8) ?
The extreme f(x)=(2x^2)/(x-8) is Maximum(0,0),Minimum(16,64)