What is the extreme f(x)=6x^{2/3}+4x ?

The extreme f(x)=6x^{2/3}+4x is Minimum(-1,2),Maximum(0,0)

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extreme f(x)=6x^{2/3}+4x

extreme points

f (x) = 6 x^{\frac{2}{3}} + 4 x

Minimum (- 1, 2), Maximum (0, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < \infty

Plug

x = - 1

into

6 x^{\frac{2}{3}} + 4 x : 2

Minimum (- 1, 2)

Plug

x = 0

into

6 x^{\frac{2}{3}} + 4 x : 0

Maximum (0, 0)

Minimum (- 1, 2), Maximum (0, 0)

f (x) = 10 - x^{4}

\frac{d}{d x} (x^{3} - x y + cos (π) (x + y))

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

f (x, y) = x + y + \frac{81}{x y}

f (x) = 2 (y + 1)^{2} + y^{2} - 2 y - 2

What is the extreme f(x)=6x^{2/3}+4x ?
The extreme f(x)=6x^{2/3}+4x is Minimum(-1,2),Maximum(0,0)