What is the extreme f(x)=(2x(x^2-6x))/((x-2)^2) ?

The extreme f(x)=(2x(x^2-6x))/((x-2)^2) is Maximum(0,0)

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extreme f(x)=(2x(x^2-6x))/((x-2)^2)

extreme points

f (x) = \frac{2 x ( x ^{2} - 6 x )}{( x - 2 ) ^{2}}

Maximum (0, 0)

Solution steps

f^{'} (x) = \frac{2 x ( x ^{2} - 6 x + 24 )}{( x - 2 ) ^{3}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

\frac{2 x ( x ^{2} - 6 x )}{( x - 2 ) ^{2}} : 0

Maximum (0, 0)

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

f (x) = \frac{( x ^{2} )}{x ^{2} - 1}

Q (x, y) = (x + 3 y) (x - 3 x y + 9 y)

y = ((x - 6)^{2} + 3)

\frac{d}{d a} (a^{5} + a^{2} b^{3} - a^{3} b^{2} - b^{5})

What is the extreme f(x)=(2x(x^2-6x))/((x-2)^2) ?
The extreme f(x)=(2x(x^2-6x))/((x-2)^2) is Maximum(0,0)