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

The extreme f(x)=(x^2)/(x-3) is Maximum(0,0),Minimum(6,12)

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

extreme points

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

Maximum (0, 0), Minimum (6, 12)

Solution steps

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

Maximum (0, 0)

Plug

x = 6

into

\frac{x ^{2}}{x - 3} : 12

Minimum (6, 12)

Maximum (0, 0), Minimum (6, 12)

1 + x

x 4 - x^{2}

\frac{2 x ^{2} + 16 x - 18}{x ^{2} + x - 6}

f (t) = \frac{3}{x - 4}

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