What is the extreme y=\sqrt[3]{x}+1/(\sqrt[3]{x)} ?

The extreme y=\sqrt[3]{x}+1/(\sqrt[3]{x)} is Maximum(-1,-2),Minimum(1,2)

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extreme y=\sqrt[3]{x}+1/(\sqrt[3]{x)}

extreme points

Maximum (- 1, - 2), Minimum (1, 2)

Solution steps

f^{'} (x) = \frac{1}{3 x ^{\frac{2}{3}}} - \frac{1}{3 x ^{\frac{4}{3}}}

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

Maximum (- 1, - 2)

Plug

x = 1

into

Minimum (1, 2)

Maximum (- 1, - 2), Minimum (1, 2)

f (x) = \frac{12 x}{3.5 x ^{2} + 2.5}

f (y) = 14 x^{2} - 2 x^{3} + 2 y^{2} + 4 x y

f (x) = x (152 - x)

f (x) = x^{2} + x y + y^{2} + 2 x - 2 y + 6

f (x) = 165000 x - 100 x^{2}

What is the extreme y=\sqrt[3]{x}+1/(\sqrt[3]{x)} ?
The extreme y=\sqrt[3]{x}+1/(\sqrt[3]{x)} is Maximum(-1,-2),Minimum(1,2)