+118658
Find f. f"(t)=9/sqrt(t), f(4)=25, f'(4)=9
\(f''(t)=9t^{-0.5}\\ f'(t)=\frac{9t^{+0.5}}{0.5}+c=18t^{0.5}+c\\ f'(4)=9\\ 18*4^{0.5}+c=9\\ 18*2+c=9\\ 36+c=9\\ c=-25\\ f'(t)=18t^{0.5}-25\\ f(t)=\frac{18t^{1.5}}{1.5}-25t+k\\ f(t)=12t^{1.5}-25t+k\\ f(4)=25\\ 12*4^{1.5}-25*4+k=25\\ 12*8-100+k=25\\ -4+k=25\\ k=29\\ f(t)=12t^{1.5}-25t+29\\\)
