Answer
$f(t)=-4t^{\frac{1}{2}}+2t+4$
Work Step by Step
$f’’(t)=t^{-\frac{3}{2}},$ $f’(4)=1,$ $f(4)=4$
$f’(t)=\int t^{-\frac{3}{2}}dt$
$f’(t)=-2t^{-\frac{1}{2}}+C_1$
Substitute in $f’(4)=1$ to solve for $C_1$
$1=-2\sqrt{\frac{1}{4}}+C_1$
$C_1=2$
$f’(t)=-2t^{-\frac{1}{2}}+2$
$f(t)=\int(-2t^{-\frac{1}{2}}+2)dt$
$f(t)=-4t^{\frac{1}{2}}+2t+C_2$
Substitute in $f(4)=4$ to solve for $C_2$
$4=-4\sqrt4+8+C_2$
$C_2=4$
$f(t)=-4t^{\frac{1}{2}}+2t+4$
