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

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

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

extreme points

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

Maximum (- 3, - \frac{9}{2}), Saddle (0, 0), Minimum (3, \frac{9}{2})

Solution steps

f^{'} (x) = \frac{x ^{4} - 9 x ^{2}}{( x ^{2} - 3 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < - 3,

Decreasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Decreasing

: 3 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

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

Maximum (- 3, - \frac{9}{2})

Plug

x = 0

into

\frac{x ^{3}}{x ^{2} - 3} : 0

Saddle (0, 0)

Plug

x = 3

into

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

Minimum (3, \frac{9}{2})

Maximum (- 3, - \frac{9}{2}), Saddle (0, 0), Minimum (3, \frac{9}{2})

f (x) = \frac{2 x + 9}{x - 1}

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

\frac{4 x}{x + 3}

- 4 x + y = - 3

g (x) = 5 (x - 2)

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