What is the critical g(t)=tsqrt(4-t) ?

The critical g(t)=tsqrt(4-t) is t= 8/3 ,t=4

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critical g(t)=tsqrt(4-t)

critical points

g (t) = t 4 - t

t = \frac{8}{3}, t = 4

Solution steps

t 4 - t

Find where

f^{'} (t)

is equal to zero or undefined

t = \frac{8}{3}, t = 4

Identify critical points not in

f (t)

domain

Domain of

t 4 - t : t \leq 4

All critical points are in domain

t = \frac{8}{3}, t = 4

f (x) = \frac{( x + \frac{1}{x} + 11 )}{x + \frac{1}{x} + 2}

f (x) = \frac{1}{4} x - \frac{1}{2}

\frac{2 e ^{x} + 3}{e ^{x} - 4}

- \frac{cos ( π \frac{11 x}{6} )}{2} - 2

f (x) = \frac{1}{9 - t}

What is the critical g(t)=tsqrt(4-t) ?
The critical g(t)=tsqrt(4-t) is t= 8/3 ,t=4