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

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

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

extreme points

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

Minimum (- \frac{4}{3}, 4)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - \frac{4}{3},

Increasing

: - \frac{4}{3} < x < \infty

Plug

x = - \frac{4}{3}

into

4 + (4 + 3 x)^{\frac{2}{3}} : 4

Minimum (- \frac{4}{3}, 4)

f (x) = 6 x + \frac{6}{x}

f (x, y) = x^{2} - 3 x y - y^{2}

p (x) = 12 x^{4} - 27 x^{3} + a x^{2} + 27 x - 6

f (x, y) = x^{3} - 3 x + y^{2} + 6 y + 8

f (x, y) = 8 + 76 x y + 38 x^{2} + 240 y + \frac{y ^{4}}{4}

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