What is the extreme f(x)=(2-3x)/(\sqrt[3]{x+2)} ?

The extreme f(x)=(2-3x)/(\sqrt[3]{x+2)} is Maximum(-10/3 ,-(12\sqrt[3]{3})/(\sqrt[3]{4)})

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extreme f(x)=(2-3x)/(\sqrt[3]{x+2)}

extreme points

Solution steps

Find intervals:

Increasing

: - \infty < x < - \frac{10}{3},

Decreasing

: - \frac{10}{3} < x < - 2,

Decreasing

: - 2 < x < \infty

Plug

x = - \frac{10}{3}

into

f (x) = \frac{5 e ^{4 x}}{5 x - 3}

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

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

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

81 - x^{2}, - 9 \leq x \leq 9

What is the extreme f(x)=(2-3x)/(\sqrt[3]{x+2)} ?
The extreme f(x)=(2-3x)/(\sqrt[3]{x+2)} is Maximum(-10/3 ,-(12\sqrt[3]{3})/(\sqrt[3]{4)})