Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. $f(x)=2 \sqrt{x}$ , $[0,25]$
A) $c=\frac{1}{5}$
B) $c=0$
C) $c=5$
D) $c=$ $\frac{25}{4}$
E) None of these

Question

Quizplus · Accepted Answer

To apply the Mean Value Theorem, we need to check three conditions:

1. $f(x)$ is continuous on $[0,25]$: $f(x) = 2\sqrt{x}$ is continuous on $[0,25]$ since it is a polynomial with a finite domain.

2. $f(x)$ is differentiable on $(0,25)$: $f(x)$ is differentiable on $(0,25)$ since it is a composition of a power function and a constant function.

3. $f'(c) = \frac{f(25) - f(0)}{25 - 0}$ for some $c \in (0,25)$: 
$f'(x) = \frac{d}{dx} (2\sqrt{x}) = \frac{1}{\sqrt{x}}$.
So, $f'(c) = \frac{1}{\sqrt{c}}$ for some $c \in (0,25)$.
Also, $f(25) = 2\sqrt{25} = 10$ and $f(0) = 2\sqrt{0} = 0$.
Therefore, $\frac{f(25) - f(0)}{25 - 0} = \frac{10}{25} = \frac{2}{5}$.
We want to find $c$ such that $f'(c) = \frac{2}{5}$.
Solving $\frac{1}{\sqrt{c}} = \frac{2}{5}$, we get $c = \frac{25}{4}$.
So the answer is D.

Find the Number C That Satisfies the Conclusion of the Mean