What is the extreme f(x)=(4x^2)/(x-8) ?

The extreme f(x)=(4x^2)/(x-8) is Maximum(0,0),Minimum(16,128)

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extreme f(x)=(4x^2)/(x-8)

extreme points

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

Maximum (0, 0), Minimum (16, 128)

Solution steps

f^{'} (x) = \frac{4 ( 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{4 x ^{2}}{x - 8} : 0

Maximum (0, 0)

Plug

x = 16

into

\frac{4 x ^{2}}{x - 8} : 128

Minimum (16, 128)

Maximum (0, 0), Minimum (16, 128)

f (x) = - 2 x^{2} + 12 x + 3

f (x, y) = - y^{2} + x y - x^{3} + y + 3

f (x, y) = 2 x^{2} + 3 y^{2} + 2 x y + 10 x - 20 y

f (x, y) = 48 x y - 32 x^{3} - 24 y^{2}

f (x, y) = x^{2} + x y + y^{2} - 34 y + 385

What is the extreme f(x)=(4x^2)/(x-8) ?
The extreme f(x)=(4x^2)/(x-8) is Maximum(0,0),Minimum(16,128)