What is the extreme f(x)=(2x^2)/(x-3) ?

The extreme f(x)=(2x^2)/(x-3) is Maximum(0,0),Minimum(6,24)

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extreme f(x)=(2x^2)/(x-3)

extreme points

f (x) = \frac{2 x ^{2}}{x - 3}

Maximum (0, 0), Minimum (6, 24)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 3,

Decreasing

: 3 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 0

into

\frac{2 x ^{2}}{x - 3} : 0

Maximum (0, 0)

Plug

x = 6

into

\frac{2 x ^{2}}{x - 3} : 24

Minimum (6, 24)

Maximum (0, 0), Minimum (6, 24)

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

f (x) = x^{3} + 2 x^{2} - 7 x, 0 \leq x \leq 2

f (x) = 2 x^{3} - 21 x^{2} + 72 x - 72.5

f (x, y) = (x + 1)^{2} + x^{2} y^{2}

f (x, y) = 5 x^{2} + 5 y^{2} + 3

What is the extreme f(x)=(2x^2)/(x-3) ?
The extreme f(x)=(2x^2)/(x-3) is Maximum(0,0),Minimum(6,24)