Question

Determine if the Mean Value Theorem for Integrals applies to the function f of x equals the square root of x minus 1 on the interval [1, 5]. If so, find the x-coordinates of the point(s) guaranteed to exist by the theorem.

The lesson didn't cover how to do this, please help! Thank you in advance.

Answer 1

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"


`\bf f(x)=√(x-1)\qquad \begin{array}{ccllll} [1&,&5]\\ a&&b \end{array}\\\\ -----------------------------\\\\ \cfrac{dy}{dx}=\cfrac{1}{2}(x-1)^{-(1)/(2)}\implies \cfrac{dy}{dx}=\cfrac{1}{2√(x-1)} \\\\\\ \textit{now, the mean value theorem says} \\\\\\ f'(c)=\cfrac{f(b)-f(a)}{b-a}\implies \cfrac{1}{2√(c-1)}=\cfrac{f(5)-f(1)}{5-1} \\\\\\ \cfrac{1}{2√(c-1)}=\cfrac{2-0}{4}\implies \cfrac{1}{2√(c-1)}=\cfrac{1}{2} \\\\\\ 1=√(c-1)\implies 1^2+1=c\implies \boxed{2=c}`