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

The extreme f(x)=2x+3\sqrt[3]{x^2} is Maximum(-1,1),Minimum(0,0)

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

extreme points

Maximum (- 1, 1), Minimum (0, 0)

Solution steps

Find the critical points:

x = - 1, x = 0

Domain of

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < \infty

Plug the extreme point

x = - 1

into

Maximum (- 1, 1)

Plug the extreme point

x = 0

into

Minimum (0, 0)

Maximum (- 1, 1), Minimum (0, 0)

f (x) = x^{3} - 4 x

x - \frac{6}{\frac{8}{x}}

\frac{7}{x}

y = 5^{(x - 3)} - 11

What is the extreme f(x)=2x+3\sqrt[3]{x^2} ?
The extreme f(x)=2x+3\sqrt[3]{x^2} is Maximum(-1,1),Minimum(0,0)