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

The extreme f(x)=x-3\sqrt[3]{x}+2 is Minimum(-1,4),Maximum(1,0)

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

extreme points

Minimum (- 1, 4), Maximum (1, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

Minimum (- 1, 4)

Plug

x = 1

into

Maximum (1, 0)

Minimum (- 1, 4), Maximum (1, 0)

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

f (x) = \frac{3 x}{2 x ^{2} + 2}, - 4 \leq x \leq 4

f (x) = cos (7 x)

f (x) = x - 2 ln (x)

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

What is the extreme f(x)=x-3\sqrt[3]{x}+2 ?
The extreme f(x)=x-3\sqrt[3]{x}+2 is Minimum(-1,4),Maximum(1,0)