find the number c that satisfies the conclusion of the mean value theorem on the given interval. (enter your…

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))

[ \begin{align*} f(4)&=\sqrt{4} = 2\ f(0)&=\sqrt{0}=0 \end{align*} ] 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)