so the graph of that, is just the graph of √(x) but with a -1 for a horizontal shift, so is √(x) but shifted to the right by 1 unit, look at the graph below
so is continuous and thus differentiable
so.. it has a point "c" whose slope if the same as the secant "ab"
[tex]\bf f(x)=\sqrt{x-1}\qquad
\begin{array}{ccllll}
[1&,&5]\\
a&&b
\end{array}\\\\
-----------------------------\\\\
\cfrac{dy}{dx}=\cfrac{1}{2}(x-1)^{-\frac{1}{2}}\implies \cfrac{dy}{dx}=\cfrac{1}{2\sqrt{x-1}}
\\\\\\
\textit{now, the mean value theorem says}
\\\\\\
f'(c)=\cfrac{f(b)-f(a)}{b-a}\implies \cfrac{1}{2\sqrt{c-1}}=\cfrac{f(5)-f(1)}{5-1}
\\\\\\
\cfrac{1}{2\sqrt{c-1}}=\cfrac{2-0}{4}\implies \cfrac{1}{2\sqrt{c-1}}=\cfrac{1}{2}
\\\\\\
1=\sqrt{c-1}\implies 1^2+1=c\implies \boxed{2=c}[/tex]
