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

The extreme y=(t-3t^2)^{1/3} is Maximum(1/6 ,(1/12)^{1/3})

extreme y=(t-3t^2)^{1/3}

extreme points

y = (t - 3 t^{2})^{\frac{1}{3}}

Maximum (\frac{1}{6}, (\frac{1}{12})^{\frac{1}{3}})

Solution steps

f^{'} (t) = \frac{1 - 6 t}{3 ( t - 3 t ^{2} ) ^{\frac{2}{3}}}

Find intervals:

Increasing

: - \infty < t < 0,

Increasing

: 0 < t < \frac{1}{6},

Decreasing

: \frac{1}{6} < t < \frac{1}{3},

Decreasing

: \frac{1}{3} < t < \infty

Plug

t = \frac{1}{6}

into

(t - 3 t^{2})^{\frac{1}{3}} : (\frac{1}{12})^{\frac{1}{3}}

Maximum (\frac{1}{6}, (\frac{1}{12})^{\frac{1}{3}})

f (x) = Y (x) = - 2 x^{3} - 3 x^{2} + 3 x + 3

f (1) = x^{3} - 6 x^{2} + 9 x + 2

f (x) = \frac{2}{3} x^{3} + \frac{11}{2} x^{2} - 21 x + 5

f (x) = (x - 3) e^{- 5 x}

f (x) = - 2^{3} - 3 \cdot (- 2^{2}) + 3 \cdot (- 2) + 1

What is the extreme y=(t-3t^2)^{1/3} ?
The extreme y=(t-3t^2)^{1/3} is Maximum(1/6 ,(1/12)^{1/3})