What is the extreme y=(2x^2)/(x-6)[-3] ?

The extreme y=(2x^2)/(x-6)[-3] is Minimum(0,0),Maximum(12,-144)

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extreme y=(2x^2)/(x-6)[-3]

extreme points

y = \frac{2 x ^{2}}{x - 6} [- 3]

Minimum (0, 0), Maximum (12, - 144)

Solution steps

f^{'} (x) = - \frac{6 ( x ^{2} - 12 x )}{( x - 6 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 6,

Increasing

: 6 < x < 12,

Decreasing

: 12 < x < \infty

Plug

x = 0

into

\frac{2 x ^{2}}{x - 6} [- 3] : 0

Minimum (0, 0)

Plug

x = 12

into

\frac{2 x ^{2}}{x - 6} [- 3] : - 144

Maximum (12, - 144)

Minimum (0, 0), Maximum (12, - 144)

f (x, y) = x - 2 y

3 x + 1 [3.6]

f (x) = \frac{9 x}{x - 4}, 6 \leq x \leq 12

f (x) = 3 x^{4} - 8 x^{3} + 6 x^{2} + 1

What is the extreme y=(2x^2)/(x-6)[-3] ?
The extreme y=(2x^2)/(x-6)[-3] is Minimum(0,0),Maximum(12,-144)