What is the extreme (2x)/(x^2+x-20)(3x)/(x^2+4x-5) ?

The extreme (2x)/(x^2+x-20)(3x)/(x^2+4x-5) is Minimum(0,0),Maximum(1.78238…,-0.23882…)

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extreme (2x)/(x^2+x-20)(3x)/(x^2+4x-5)

extreme points

\frac{2 x}{x ^{2} + x - 20} \frac{3 x}{x ^{2} + 4 x - 5}

Minimum (0, 0), Maximum (1.78238 \dots, - 0.23882 \dots)

Solution steps

Find the critical points:

x = 0, x \approx 1.78238 \dots

Domain of

\frac{2 x}{x ^{2} + x - 20} \frac{3 x}{x ^{2} + 4 x - 5} : x < - 5 or - 5 < x < 1 or 1 < x < 4 or x > 4

Find intervals:

Increasing

: - \infty < x < - 5,

Decreasing

: - 5 < x < 0,

Increasing

: 0 < x < 1,

Increasing

: 1 < x < 1.78238 \dots,

Decreasing

: 1.78238 \dots < x < 4,

Decreasing

: 4 < x < \infty

Plug the extreme point

x = 0

into

\frac{2 x}{x ^{2} + x - 20} \frac{3 x}{x ^{2} + 4 x - 5} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 1.78238 \dots

into

\frac{2 x}{x ^{2} + x - 20} \frac{3 x}{x ^{2} + 4 x - 5} \Rightarrow y = - 0.23882 \dots

Maximum (1.78238 \dots, - 0.23882 \dots)

Minimum (0, 0), Maximum (1.78238 \dots, - 0.23882 \dots)

f (x) = 2 x^{3} - 12 x^{2} + 14

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

f (x) = e^{x^{3} - 27 x}

\frac{x ^{2} - 11 x + 67}{x - 9}

f (x) = 6 x^{\frac{2}{3}} - x

What is the extreme (2x)/(x^2+x-20)(3x)/(x^2+4x-5) ?
The extreme (2x)/(x^2+x-20)(3x)/(x^2+4x-5) is Minimum(0,0),Maximum(1.78238…,-0.23882…)