## Calculus 10th Edition

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