What is the extreme f(x)=(x^2)/(x-9) ?

The extreme f(x)=(x^2)/(x-9) is Maximum(0,0),Minimum(18,36)

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extreme f(x)=(x^2)/(x-9)

extreme points

f (x) = \frac{x ^{2}}{x - 9}

Maximum (0, 0), Minimum (18, 36)

Solution steps

f^{'} (x) = \frac{x ^{2} - 18 x}{( x - 9 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 9,

Decreasing

: 9 < x < 18,

Increasing

: 18 < x < \infty

Plug

x = 0

into

\frac{x ^{2}}{x - 9} : 0

Maximum (0, 0)

Plug

x = 18

into

\frac{x ^{2}}{x - 9} : 36

Minimum (18, 36)

Maximum (0, 0), Minimum (18, 36)

f (x) = \frac{x ^{3} - 4 x}{3 x ^{2} + 3 x - 6}

f (x) = 100 - 3 x

e^{x + 1}

arcsin (x)

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

What is the extreme f(x)=(x^2)/(x-9) ?
The extreme f(x)=(x^2)/(x-9) is Maximum(0,0),Minimum(18,36)