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

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)

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)