Question

Please Help Let f(t)=t(Sqrt4-t) on the interval [-1,3]. Find the absolute maximum and absolute minimum.

Answer 1

\[f \left( t \right)=t \sqrt{4-t}\]
\[f'\left( t \right)=t \frac{ -1 }{ 2\sqrt{4-t} }+\sqrt{4-t}*1=\frac{ -t+2\left( 4-t \right) }{ 2\sqrt{4-t} } =\frac{ 8-3t }{ 2\sqrt{4-t} }\]
f'(t)=0,gives 8-3t=0,\[t=\frac{ 8 }{ 3 }\]
plug t=-1,3,8/3 in f(x) and calculate absolute maximum and absolute minimum.

Answer 2

thanx

Answer 3

YW