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
$$ f'(c)= \displaystyle{ f(4)-f(0) \over 4-0 } \ \ \ \ \longrightarrow $$
$$ \displaystyle{ 1 \over 2 \ \sqrt{c} } = \displaystyle{ (3 + \sqrt{4} \ ) - (3 + \sqrt{0} \ ) \over 4-0 } \ \ \ \ \longrightarrow $$
$$ \displaystyle{ 1 \over 2 \ \sqrt{c} } = \displaystyle{ 5 - 3 \over 4 } \ \ \ \ \longrightarrow $$
$$ \displaystyle{ 1 \over 2 \ \sqrt{c} } = \displaystyle{ 1 \over 2 } \ \ \ \ \longrightarrow $$
$$ \sqrt{c} = 1 \ \ \ \ \longrightarrow $$
$$ (\sqrt{c} \ )^2 = 1^2 \ \ \ \ \longrightarrow $$
$$ c = 1 $$
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