What is the extreme f(x)=(8(x^2-1))/(3x^{1/3)} ?

The extreme f(x)=(8(x^2-1))/(3x^{1/3)} is None

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extreme f(x)=(8(x^2-1))/(3x^{1/3)}

extreme points

f (x) = \frac{8 ( x ^{2} - 1 )}{3 x ^{\frac{1}{3}}}

None

Solution steps

f^{'} (x) = \frac{8 ( 5 x ^{2} + 1 )}{9 x ^{\frac{4}{3}}}

Find intervals:

None

None

f (x) = x 13 - x^{2}

x^{4} - 8 x^{2} - 10

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

f (r, h) = 2 π r h + 2 π r^{2}

g (t) = u (t)

What is the extreme f(x)=(8(x^2-1))/(3x^{1/3)} ?
The extreme f(x)=(8(x^2-1))/(3x^{1/3)} is None