> 
> Show $\sqrt{65}-8=\frac{1}{2\sqrt{c}}$ for some $c \in (64,65)$, and hence show: 
>  $$8+\frac{1}{18}<\sqrt{65}<8+\frac{1}{16}$$
> 
> 
> 

I managed to do the first part easily, but I don't know how to do the "hence show" part. 

*Working For the First Part:*

Consider the function $f(x)=x^{\frac{1}{2}}$ on $[64,65]$. Then f is continuous on $[64,65]$ and differentiable on $(64,65)$. By the mean value theorem, there is a $c \in (64,65)$ such that:
$$f'(c)=\frac{f(65)-f(64)}{65-64}$$
$f'(x)=\frac{1}{2\sqrt{x}}$, so we have:
$$\frac{1}{2\sqrt{c}}=\sqrt{65}-8$$