#### Answer

Critical Number is $\frac{8}{3}$.

#### Work Step by Step

Begin by taking the derivative of g(x)
$g(x)= t\sqrt {4-t} $, $t<3$
$g'(x)= (\sqrt {4-t} )(1) + (t)(\frac{1}{2}(4-t)^{-\frac{1}{2}})(-1) $
Simplify
$g'(x)= (4-t)^{\frac{1}{2}} - \frac{t}{2\sqrt {4-t}}$
$g'(x)= (4-t)^{\frac{1}{2}}(1-\frac{t}{2(4-t)}) $
Now set the derivative equal to zero to find the critical numbers
$0=(4-t)^{\frac{1}{2}}$
$t=4$, but since $t<3$ it cannot be a critical number
$0=(1-\frac{t}{2(4-t)})$
Simplify
$t=\frac{8}{3}$
Because $\frac{8}{3} $<3. It is a critical number