Answer
$f(x)=-4x^{\frac{1}{2}}+3x$
Work Step by Step
$f''(x)=x^{-\frac{3}{2}}\hspace{8mm}f'(4)=2\hspace{8mm}f(0)=0$
$\int f''(x)dx=\int x^{-\frac{3}{2}}dx=-2x^{-\frac{1}{2}}+C$
$f'(4)=-2\times(4)^{-\frac{1}{2}}+C=-1+C=2$
$C=3$
$f'(x)=-2x^{-\frac{1}{2}}+3$
$\int f'(x)dx=\int (-2x^{-\frac{1}{2}}+3)dx=-4x^{\frac{1}{2}}+3x+C=f(x)$
$f(0)=0=C$
$f(x)=-4x^{\frac{1}{2}}+3x$
