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

The extreme (x^2+1)/(x^3-1) is Minimum(-0.59607…,-1.11843…),Maximum(0,-1)

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extreme (x^2+1)/(x^3-1)

extreme points

\frac{x ^{2} + 1}{x ^{3} - 1}

Minimum (- 0.59607 \dots, - 1.11843 \dots), Maximum (0, - 1)

Solution steps

Find the critical points:

x \approx - 0.59607 \dots, x = 0

Domain of

\frac{x ^{2} + 1}{x ^{3} - 1} : x < 1 or x > 1

Find intervals:

Decreasing

: - \infty < x < - 0.59607 \dots,

Increasing

: - 0.59607 \dots < x < 0,

Decreasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug the extreme point

x = - 0.59607 \dots

into

\frac{x ^{2} + 1}{x ^{3} - 1} \Rightarrow y = - 1.11843 \dots

Minimum (- 0.59607 \dots, - 1.11843 \dots)

Plug the extreme point

x = 0

into

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

Maximum (0, - 1)

Minimum (- 0.59607 \dots, - 1.11843 \dots), Maximum (0, - 1)

f (x) = - 17 \frac{x}{x ^{2} + y ^{2} + 1}

f (x) = x e^{- \frac{x}{4}}

f (x, y) = 350 \cdot x + 200 \cdot y

- x^{2} + 100 x + 8

What is the extreme (x^2+1)/(x^3-1) ?
The extreme (x^2+1)/(x^3-1) is Minimum(-0.59607…,-1.11843…),Maximum(0,-1)