What is the extreme (-x^3-x+5)/(2x^3+3x^2-7) ?

The extreme (-x^3-x+5)/(2x^3+3x^2-7) is Minimum(-1.64948…,-1.42541…),Maximum(0.22450…,-0.69792…)

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extreme (-x^3-x+5)/(2x^3+3x^2-7)

extreme points

\frac{- x ^{3} - x + 5}{2 x ^{3} + 3 x ^{2} - 7}

Minimum (- 1.64948 \dots, - 1.42541 \dots), Maximum (0.22450 \dots, - 0.69792 \dots)

Solution steps

Find the critical points:

x \approx - 1.64948 \dots, x \approx 0.22450 \dots

Domain of

\frac{- x ^{3} - x + 5}{2 x ^{3} + 3 x ^{2} - 7} : x < 1.14937 \dots or x > 1.14937 \dots

Find intervals:

Decreasing

: - \infty < x < - 1.64948 \dots,

Increasing

: - 1.64948 \dots < x < 0.22450 \dots,

Decreasing

: 0.22450 \dots < x < 1.14937 \dots,

Decreasing

: 1.14937 \dots < x < \infty

Plug the extreme point

x = - 1.64948 \dots

into

\frac{- x ^{3} - x + 5}{2 x ^{3} + 3 x ^{2} - 7} \Rightarrow y = - 1.42541 \dots

Minimum (- 1.64948 \dots, - 1.42541 \dots)

Plug the extreme point

x = 0.22450 \dots

into

\frac{- x ^{3} - x + 5}{2 x ^{3} + 3 x ^{2} - 7} \Rightarrow y = - 0.69792 \dots

Maximum (0.22450 \dots, - 0.69792 \dots)

Minimum (- 1.64948 \dots, - 1.42541 \dots), Maximum (0.22450 \dots, - 0.69792 \dots)

y = \frac{x}{2} + \frac{2}{x}

f (x) = \frac{2 x - 2}{x ^{2} - 10 x + 25}

f (x, y) = 5 x y

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

\frac{x ^{2} - x - 6}{2 x + 4}

What is the extreme (-x^3-x+5)/(2x^3+3x^2-7) ?
The extreme (-x^3-x+5)/(2x^3+3x^2-7) is Minimum(-1.64948…,-1.42541…),Maximum(0.22450…,-0.69792…)