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
$$ f'(c)= \displaystyle{ f(5)-f(1) \over 5 - 1 } \ \ \ \ \longrightarrow $$
$$ 2 + \displaystyle{ 1 \over 2 \sqrt{c-1} } = \displaystyle{ (2(5) + \sqrt{5-1} \ ) - (2(1) + \sqrt{1-1} \ ) \over 4 } \ \ \ \ \longrightarrow $$
$$ 2 + \displaystyle{ 1 \over 2 \sqrt{c-1} } = \displaystyle{ (10 + 2 ) - ( 2 + 0) \over 4 } \ \ \ \ \longrightarrow $$
$$ 2 + \displaystyle{ 1 \over 2 \sqrt{c-1} } = \displaystyle{ 5 \over 2 } \ \ \ \ \longrightarrow $$
$$ \displaystyle{ 1 \over 2 \sqrt{c-1} } = \displaystyle{ 1 \over 2 } \ \ \ \ \longrightarrow $$
$$ \sqrt{c-1} = 1 \ \ \ \ \longrightarrow $$
$$ (\sqrt{c-1} \ )^2 = 1^2 \ \ \ \ \longrightarrow $$
$$ c-1 = 1 \ \ \ \ \longrightarrow $$
$$ c = 2 $$
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