What is the extreme (10x^3-2)/(2x^3+5x^2+9x) ?

The extreme (10x^3-2)/(2x^3+5x^2+9x) is Maximum(-3.55613…,7.69301…),Minimum(-0.41929…,0.89977…)

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extreme (10x^3-2)/(2x^3+5x^2+9x)

extreme points

\frac{10 x ^{3} - 2}{2 x ^{3} + 5 x ^{2} + 9 x}

Maximum (- 3.55613 \dots, 7.69301 \dots), Minimum (- 0.41929 \dots, 0.89977 \dots)

Solution steps

Find the critical points:

x \approx - 3.55613 \dots, x \approx - 0.41929 \dots

Domain of

\frac{10 x ^{3} - 2}{2 x ^{3} + 5 x ^{2} + 9 x} : x < 0 or x > 0

Find intervals:

Increasing

: - \infty < x < - 3.55613 \dots,

Decreasing

: - 3.55613 \dots < x < - 0.41929 \dots,

Increasing

: - 0.41929 \dots < x < 0,

Increasing

: 0 < x < \infty

Plug the extreme point

x = - 3.55613 \dots

into

\frac{10 x ^{3} - 2}{2 x ^{3} + 5 x ^{2} + 9 x} \Rightarrow y = 7.69301 \dots

Maximum (- 3.55613 \dots, 7.69301 \dots)

Plug the extreme point

x = - 0.41929 \dots

into

\frac{10 x ^{3} - 2}{2 x ^{3} + 5 x ^{2} + 9 x} \Rightarrow y = 0.89977 \dots

Minimum (- 0.41929 \dots, 0.89977 \dots)

Maximum (- 3.55613 \dots, 7.69301 \dots), Minimum (- 0.41929 \dots, 0.89977 \dots)

- 6 x^{2} + 336 x - 4320

f (x) = \frac{12 x}{3.5 x ^{2} + 2.5}

f (y) = 14 x^{2} - 2 x^{3} + 2 y^{2} + 4 x y

f (x) = x (152 - x)

What is the extreme (10x^3-2)/(2x^3+5x^2+9x) ?
The extreme (10x^3-2)/(2x^3+5x^2+9x) is Maximum(-3.55613…,7.69301…),Minimum(-0.41929…,0.89977…)