SOLUTION 1: We are given the function $f(x) = 3 + \sqrt{x}$ and the interval $[0, 4]$. This function is continuous on the closed interval $[0, 4]$ since it is the sum of the two continuous functions $y=3$ and $y= \sqrt{x}$. The derivative of $f$ is $$f'(x) = 0 + (1/2) x^{-1/2} = \displaystyle{ 1 \over 2 \ \sqrt{x} }$$ We can now see that $f$ is differentiable on the open interval $(0, 4)$. 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 $(0, 4)$. Then $$f'(c)= \displaystyle{ f(4)-f(0) \over 4-0 } \ \ \ \ \longrightarrow$$ $$\displaystyle{ 1 \over 2 \ \sqrt{c} } = \displaystyle{ (3 + \sqrt{4} \ ) - (3 + \sqrt{0} \ ) \over 4-0 } \ \ \ \ \longrightarrow$$ $$\displaystyle{ 1 \over 2 \ \sqrt{c} } = \displaystyle{ 5 - 3 \over 4 } \ \ \ \ \longrightarrow$$ $$\displaystyle{ 1 \over 2 \ \sqrt{c} } = \displaystyle{ 1 \over 2 } \ \ \ \ \longrightarrow$$ $$\sqrt{c} = 1 \ \ \ \ \longrightarrow$$ $$(\sqrt{c} \ )^2 = 1^2 \ \ \ \ \longrightarrow$$ $$c = 1$$