What is the extreme f(x)=(8-9x)/(\sqrt[3]{x+3)} ?

The extreme f(x)=(8-9x)/(\sqrt[3]{x+3)} is Maximum(-4.94444…,-42.06240…)

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extreme f(x)=(8-9x)/(\sqrt[3]{x+3)}

extreme points

Maximum (- 4.94444 \dots, - 42.06240 \dots)

Solution steps

Find the critical points:

x \approx - 4.94444 \dots

Domain of

Find intervals:

Increasing

: - \infty < x < - 4.94444 \dots,

Decreasing

: - 4.94444 \dots < x < - 3,

Decreasing

: - 3 < x < \infty

Plug the extreme point

x = - 4.94444 \dots

into

Maximum (- 4.94444 \dots, - 42.06240 \dots)

f (x, y) = 2 x^{3} + 2 x y^{2} - 4 x + 5

f (x) = \frac{40}{r} + 4 π r^{2}

f (x) = x \cdot x + 2

f (x, y) = 13 - 4 x + 8 y

g (x) = x + \frac{4}{x ^{2}}

What is the extreme f(x)=(8-9x)/(\sqrt[3]{x+3)} ?
The extreme f(x)=(8-9x)/(\sqrt[3]{x+3)} is Maximum(-4.94444…,-42.06240…)