Answer
$f''(0)=\dfrac{3}{128}.$
Work Step by Step
$f(x)=(x+4)^{-\frac{1}{2}}$
$f'(x)=-\dfrac{1}{2}(x+4)^{-\frac{1}{2}-1}=-\dfrac{1}{2\sqrt{(x+4)^3}}$
$f'(x)=-\dfrac{1}{2}(x+4)^-\frac{3}{2}$
$f''(x)=(-\dfrac{1}{2})(-\dfrac{3}{2})(x+4)^{-\frac{3}{2}-1}=\dfrac{3}{4\sqrt{(x+4)^5}}$
$f''(0)=\dfrac{3}{4\sqrt{(0+4)^5}}=\dfrac{3}{128}.$
