What is the extreme f(x)= 1/(x(12-x^2)) ?

The extreme f(x)= 1/(x(12-x^2)) is Maximum(-2,-1/16),Minimum(2, 1/16)

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extreme f(x)= 1/(x(12-x^2))

extreme points

f (x) = \frac{1}{x ( 12 - x ^{2} )}

Maximum (- 2, - \frac{1}{16}), Minimum (2, \frac{1}{16})

Solution steps

f^{'} (x) = - \frac{12 - 3 x ^{2}}{x ^{2} ( 12 - x ^{2} ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 23,

Increasing

: - 23 < x < - 2,

Decreasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < 23,

Increasing

: 23 < x < \infty

Plug

x = - 2

into

\frac{1}{x ( 12 - x ^{2} )} : - \frac{1}{16}

Maximum (- 2, - \frac{1}{16})

Plug

x = 2

into

\frac{1}{x ( 12 - x ^{2} )} : \frac{1}{16}

Minimum (2, \frac{1}{16})

Maximum (- 2, - \frac{1}{16}), Minimum (2, \frac{1}{16})

What is the extreme f(x)= 1/(x(12-x^2)) ?
The extreme f(x)= 1/(x(12-x^2)) is Maximum(-2,-1/16),Minimum(2, 1/16)