Question

find the number c that satisfies the conclusion of the mean value theorem on the given interval. (enter your answers as a comma - separated list. if an answer does not exist, enter dne.)
$f(x)=\sqrt{x}$, $0,4$

Explanation:

Step1: Check the conditions of the Mean Value Theorem

The function \( f(x)=\sqrt{x}=x^{\frac{1}{2}} \) is continuous on \([0,4]\) and differentiable on \((0,4)\) since \( f^{\prime}(x)=\frac{1}{2\sqrt{x}}\) exists for \(x\in(0,4)\).

Step2: Calculate \(f(4)\) and \(f(0)\)

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The slope of the secant line is \(\frac{f(4)-f(0)}{4 - 0}=\frac{2-0}{4}=\frac{1}{2}\)

Step3: Find \(f^{\prime}(c)\) and solve for \(c\)

We know that \(f^{\prime}(x)=\frac{1}{2\sqrt{x}}\), and by the Mean Value Theorem \(f^{\prime}(c)=\frac{f(4)-f(0)}{4 - 0}\)
Set \(\frac{1}{2\sqrt{c}}=\frac{1}{2}\)
Cross - multiply gives \(2 = 2\sqrt{c}\), then \(\sqrt{c}=1\), so \(c = 1\)

Answer:

\(1\)