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

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

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < - 3,

Increasing

: - 3 < x < 0,

Increasing

: 0 < x < 3,

Increasing

: 3 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = - 3

into

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

Minimum (- 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}

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

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

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