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

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

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

extreme points

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

Maximum (0, 0), Minimum (4, 8)

Solution steps

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

Maximum (0, 0)

Plug

x = 4

into

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

Minimum (4, 8)

Maximum (0, 0), Minimum (4, 8)

x^{4} - x^{3} - 3 x^{2} + 5 x - 2

2 x^{3} + 9 x y^{2} + 15 x^{2} + 27 y^{2}

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

g (x, y) = x y^{2} + x^{2} + y

K (r, s) = 6 r - s

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