What is the extreme f(x)=(t-4t^2)^{1/3} ?

The extreme f(x)=(t-4t^2)^{1/3} is Maximum(1/8 , 1/(2*2^{1/3)})

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extreme f(x)=(t-4t^2)^{1/3}

extreme

Maximum (\frac{1}{8}, \frac{1}{2 \cdot 2 ^{\frac{1}{3}}})

Solution steps

f^{'} (t) = \frac{1 - 8 t}{3 ( t - 4 t ^{2} ) ^{\frac{2}{3}}}

Find intervals:

Increasing

: - \infty < t < 0,

Increasing

: 0 < t < \frac{1}{8},

Decreasing

: \frac{1}{8} < t < \frac{1}{4},

Decreasing

: \frac{1}{4} < t < \infty

Plug

t = \frac{1}{8}

into

(t - 4 t^{2})^{\frac{1}{3}} : \frac{1}{2 \cdot 2 ^{\frac{1}{3}}}

Maximum (\frac{1}{8}, \frac{1}{2 \cdot 2 ^{\frac{1}{3}}})

What is the extreme f(x)=(t-4t^2)^{1/3} ?
The extreme f(x)=(t-4t^2)^{1/3} is Maximum(1/8 , 1/(2*2^{1/3)})