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

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

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

extreme points

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

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

Solution steps

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

Maximum (0, 0)

Plug

x = 8

into

\frac{4 x ^{2}}{2 x - 8} : 32

Minimum (8, 32)

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

f (x, y) = x + \frac{4}{x ^{2} y} + y

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

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

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

y = 2^{x} \cdot I n (5)

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