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

The extreme f(x)=(3x^2)/(x-3) is Maximum(0,0),Minimum(6,36)

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

extreme points

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

Maximum (0, 0), Minimum (6, 36)

Solution steps

f^{'} (x) = \frac{3 ( 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{3 x ^{2}}{x - 3} : 0

Maximum (0, 0)

Plug

x = 6

into

\frac{3 x ^{2}}{x - 3} : 36

Minimum (6, 36)

Maximum (0, 0), Minimum (6, 36)

(010), (02)

f (x) = \frac{x}{( 8 x + 1 )}

f (x) = - 1 - x^{5}

e^{2 x + 1}

- 4 x - 14

What is the extreme f(x)=(3x^2)/(x-3) ?
The extreme f(x)=(3x^2)/(x-3) is Maximum(0,0),Minimum(6,36)