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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < \infty

Plug

x = - 1

into

2 x + 3 x^{\frac{2}{3}} : 1

Minimum (- 1, 1)

Plug

x = 0

into

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

Maximum (0, 0)

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

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

\frac{x ^{2} - 4}{x + 2}

f (x) = 4 \frac{7 - x}{x}

y - 4 = \frac{1}{6} (x - 6)

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

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