What is the extreme f(x)=(t^2)/(1+t^3) ?

The extreme f(x)=(t^2)/(1+t^3) is Minimum(0,0),Maximum(\sqrt[3]{2},(2^{2/3})/3)

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extreme f(x)=(t^2)/(1+t^3)

extreme

Minimum (0, 0), Maximum (32, \frac{2 ^{\frac{2}{3}}}{3})

Solution steps

f^{'} (t) = \frac{2 t - t ^{4}}{( 1 + t ^{3} ) ^{2}}

Find intervals:

Decreasing

: - \infty < t < - 1,

Decreasing

: - 1 < t < 0,

Increasing

: 0 < t < 32,

Decreasing

: 32 < t < \infty

Plug

t = 0

into

\frac{t ^{2}}{1 + t ^{3}} : 0

Minimum (0, 0)

Plug

t = 32

into

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

Maximum (32, \frac{2 ^{\frac{2}{3}}}{3})

Minimum (0, 0), Maximum (32, \frac{2 ^{\frac{2}{3}}}{3})

What is the extreme f(x)=(t^2)/(1+t^3) ?
The extreme f(x)=(t^2)/(1+t^3) is Minimum(0,0),Maximum(\sqrt[3]{2},(2^{2/3})/3)