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

The extreme f(x)=x^4+4/(3x^3) is Minimum(1, 7/3)

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extreme f(x)=x^4+4/(3x^3)

extreme

Minimum (1, \frac{7}{3})

Solution steps

Find the critical points:

x = 1

Domain of

x^{4} + \frac{4}{3 x ^{3}} : x < 0 or x > 0

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug the extreme point

x = 1

into

x^{4} + \frac{4}{3 x ^{3}} \Rightarrow y = \frac{7}{3}

Minimum (1, \frac{7}{3})

What is the extreme f(x)=x^4+4/(3x^3) ?
The extreme f(x)=x^4+4/(3x^3) is Minimum(1, 7/3)