What is the extreme-(40/3)/(x^2)+0.54x ?

The extreme-(40/3)/(x^2)+0.54x is Maximum(-\sqrt[3]{(6.17284E15)/(1.25E14)},-2.97173…)

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extreme-(40/3)/(x^2)+0.54x

extreme

Maximum (- 3 \frac{6.17284 E 15}{1.25 E 14}, - 2.97173 \dots)

Solution steps

Find the critical points:

x = - 3 \frac{6.17284 E 15}{1.25 E 14}

Domain of

- \frac{\frac{40}{3}}{x ^{2}} + 0.54 x : x < 0 or x > 0

Find intervals:

Increasing

: - \infty < x < - 3 \frac{6.17284 E 15}{1.25 E 14},

Decreasing

: - 3 \frac{6.17284 E 15}{1.25 E 14} < x < 0,

Increasing

: 0 < x < \infty

Plug the extreme point

x = - 3 \frac{6.17284 E 15}{1.25 E 14}

into

- \frac{\frac{40}{3}}{x ^{2}} + 0.54 x \Rightarrow y = - 2.97173 \dots

Maximum (- 3 \frac{6.17284 E 15}{1.25 E 14}, - 2.97173 \dots)

What is the extreme-(40/3)/(x^2)+0.54x ?
The extreme-(40/3)/(x^2)+0.54x is Maximum(-\sqrt[3]{(6.17284E15)/(1.25E14)},-2.97173…)