SOLUTION 7: We are given the function $f(x) = 2x + \sqrt{x-1}$ and the interval $[1, 5]$. This function is continuous on the closed interval $[1, 5]$ since it is the sum of continuous functions $y=2x$ (polynomial) and $y= \sqrt{x-1}$ $\Big($functional composition of continuous functions $y=x-1$ (polynomial) and $y= \sqrt{x}$ (well-known). $\Big)$. The derivative of $f$ is $$f'(x) = 2 + (1/2)(x-1)^{-1/2} = 2 + \displaystyle{ 1 \over 2\sqrt{x-1} }$$ We can now see that $f$ is differentiable on the open interval $(1, 5)$. The assumptions of the Mean Value Theorem have now been met. Let's apply the Mean Value Theorem and find all possible values of $c$ in the open interval $(1, 5)$. Then $$f'(c)= \displaystyle{ f(5)-f(1) \over 5 - 1 } \ \ \ \ \longrightarrow$$ $$2 + \displaystyle{ 1 \over 2 \sqrt{c-1} } = \displaystyle{ (2(5) + \sqrt{5-1} \ ) - (2(1) + \sqrt{1-1} \ ) \over 4 } \ \ \ \ \longrightarrow$$ $$2 + \displaystyle{ 1 \over 2 \sqrt{c-1} } = \displaystyle{ (10 + 2 ) - ( 2 + 0) \over 4 } \ \ \ \ \longrightarrow$$ $$2 + \displaystyle{ 1 \over 2 \sqrt{c-1} } = \displaystyle{ 5 \over 2 } \ \ \ \ \longrightarrow$$ $$\displaystyle{ 1 \over 2 \sqrt{c-1} } = \displaystyle{ 1 \over 2 } \ \ \ \ \longrightarrow$$ $$\sqrt{c-1} = 1 \ \ \ \ \longrightarrow$$ $$(\sqrt{c-1} \ )^2 = 1^2 \ \ \ \ \longrightarrow$$ $$c-1 = 1 \ \ \ \ \longrightarrow$$ $$c = 2$$