What is the extreme f(x)=tsqrt(4-t) ?

The extreme f(x)=tsqrt(4-t) is Maximum(8/3 ,(16)/(3sqrt(3))),Minimum(4,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=tsqrt(4-t)

extreme points

f (x) = t 4 - t

Maximum (\frac{8}{3}, \frac{16}{3 3}), Minimum (4, 0)

Solution steps

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

Find intervals:

Increasing

: - \infty < t < \frac{8}{3},

Decreasing

: \frac{8}{3} < t < 4

Plug

t = \frac{8}{3}

into

t 4 - t : \frac{16}{3 3}

Maximum (\frac{8}{3}, \frac{16}{3 3})

Plug

t = 4

into

t 4 - t : 0

Minimum (4, 0)

Maximum (\frac{8}{3}, \frac{16}{3 3}), Minimum (4, 0)

What is the extreme f(x)=tsqrt(4-t) ?
The extreme f(x)=tsqrt(4-t) is Maximum(8/3 ,(16)/(3sqrt(3))),Minimum(4,0)