What is the extreme f(x)=(x^3-27)/(x^2-9) ?

The extreme f(x)=(x^3-27)/(x^2-9) is Maximum(-6,-9),Minimum(0,3)

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extreme f(x)=(x^3-27)/(x^2-9)

extreme points

f (x) = \frac{x ^{3} - 27}{x ^{2} - 9}

Maximum (- 6, - 9), Minimum (0, 3)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 6,

Decreasing

: - 6 < x < - 3,

Decreasing

: - 3 < x < 0,

Increasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 6

into

\frac{x ^{3} - 27}{x ^{2} - 9} : - 9

Maximum (- 6, - 9)

Plug

x = 0

into

\frac{x ^{3} - 27}{x ^{2} - 9} : 3

Minimum (0, 3)

Maximum (- 6, - 9), Minimum (0, 3)

f (x, y) = 3 x^{2} + 6 y^{2}

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

f (x, y) = ln (36 - 4 x^{2} - 9 y^{2})

f (x) = \frac{1}{3} x^{3} - x^{2} + 3

f (x) = \frac{1}{3} x^{3} - x^{2} + 1

What is the extreme f(x)=(x^3-27)/(x^2-9) ?
The extreme f(x)=(x^3-27)/(x^2-9) is Maximum(-6,-9),Minimum(0,3)