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 points

- \frac{\frac{40}{3}}{x ^{2}} + 0.54 x

Solution steps

Find the critical points:

Domain of

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

Find intervals:

Increasing

Decreasing

Increasing

: 0 < x < \infty

Plug the extreme point

into

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

f (x, y) = x \cdot e^{y}

f (x) = 4 θ - 6 sin (θ)

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

f (x) = 2 x^{3} + 9 x^{2} - 24 x - 10

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

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…)