What is the extreme f(x)=sqrt(t)(1-t) ?

The extreme f(x)=sqrt(t)(1-t) is Minimum(0,0),Maximum(1/3 , 2/(3sqrt(3)))

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extreme f(x)=sqrt(t)(1-t)

extreme points

f (x) = t (1 - t)

Minimum (0, 0), Maximum (\frac{1}{3}, \frac{2}{3 3})

Solution steps

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

Find intervals:

Increasing

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

Decreasing

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

Plug

t = 0

into

t (1 - t) : 0

Minimum (0, 0)

Plug

t = \frac{1}{3}

into

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

Maximum (\frac{1}{3}, \frac{2}{3 3})

Minimum (0, 0), Maximum (\frac{1}{3}, \frac{2}{3 3})

f (x, y) = x ln (y - x)

- 4.9 x^{2} + 24 x + 8

f (x, y) = x^{3} + 3 x y^{2} - 3 x^{2} - 3 y^{2} + 10

f (x) = - \frac{2}{3} x^{3} - \frac{11}{2} x^{2} - 5 x + 1

f (x, y) = x^{3} + y^{3} - 3 x - 12 y + 13 e

What is the extreme f(x)=sqrt(t)(1-t) ?
The extreme f(x)=sqrt(t)(1-t) is Minimum(0,0),Maximum(1/3 , 2/(3sqrt(3)))