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

Explanation:

Step1: Recall Mean - Value Theorem

The Mean - Value Theorem states that if $y = f(x)$ is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then there exists at least one number $c\in(a,b)$ such that $f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}$. Here, $a = 0$, $b = 25$, and $f(x)=\sqrt{x}=x^{\frac{1}{2}}$.

Step2: Calculate $f(a)$ and $f(b)$

First, find $f(0)$ and $f(25)$. Since $f(x)=\sqrt{x}$, then $f(0)=\sqrt{0}=0$ and $f(25)=\sqrt{25}=5$. So, $\frac{f(25)-f(0)}{25 - 0}=\frac{5 - 0}{25}=\frac{1}{5}$.

Step3: Find the derivative of $f(x)$

Differentiate $f(x)=x^{\frac{1}{2}}$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$. Then $f^{\prime}(x)=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}$.

Step4: Solve for $c$

Set $f^{\prime}(c)=\frac{1}{5}$. We have $\frac{1}{2\sqrt{c}}=\frac{1}{5}$. Cross - multiply to get $2\sqrt{c}=5$. Then $\sqrt{c}=\frac{5}{2}$, and square both sides to find $c=\frac{25}{4}=6.25$.

Answer:

$6.25$