Mean Value Theorem Solution 1

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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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