What is the extreme f(x)= 1/(3x^3-9x+2) ?

The extreme f(x)= 1/(3x^3-9x+2) is Minimum(-1, 1/8),Maximum(1,-1/4)

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extreme f(x)= 1/(3x^3-9x+2)

extreme points

f (x) = \frac{1}{3 x ^{3} - 9 x + 2}

Minimum (- 1, \frac{1}{8}), Maximum (1, - \frac{1}{4})

Solution steps

Find the critical points:

x = - 1, x = 1

Domain of

\frac{1}{3 x ^{3} - 9 x + 2} : x < - 1.83398 \dots or - 1.83398 \dots < x < 0.22607 \dots or 0.22607 \dots < x < 1.60791 \dots or x > 1.60791 \dots

Find intervals:

Decreasing

: - \infty < x < - 1.83398 \dots,

Decreasing

: - 1.83398 \dots < x < - 1,

Increasing

: - 1 < x < 0.22607 \dots,

Increasing

: 0.22607 \dots < x < 1,

Decreasing

: 1 < x < 1.60791 \dots,

Decreasing

: 1.60791 \dots < x < \infty

Plug the extreme point

x = - 1

into

\frac{1}{3 x ^{3} - 9 x + 2} \Rightarrow y = \frac{1}{8}

Minimum (- 1, \frac{1}{8})

Plug the extreme point

x = 1

into

\frac{1}{3 x ^{3} - 9 x + 2} \Rightarrow y = - \frac{1}{4}

Maximum (1, - \frac{1}{4})

Minimum (- 1, \frac{1}{8}), Maximum (1, - \frac{1}{4})

y = \frac{( \frac{1}{x - 1} - \frac{2}{x} + z )}{2 + \frac{1}{x}}

f (x) = x^{4} - 8 x^{2} + 7

f (x) = - 2 x^{3} + 45 x^{2} - 300 x

f (x) = - x^{2} - 4 x + 5

x 2 - x^{2}

What is the extreme f(x)= 1/(3x^3-9x+2) ?
The extreme f(x)= 1/(3x^3-9x+2) is Minimum(-1, 1/8),Maximum(1,-1/4)