To solve the problem using the Mean Value Theorem (MVT) for the function $y(x) = \sqrt{x}$ on the interval $[25, 289]$, we follow these steps:

Mean Value Theorem Statement

The MVT states that for a function $f(x)$ continuous on $[a, b]$ and differentiable on $(a, b)$, there exists a point $c \in (a, b)$ such that: $f'(c) = \frac{f(b) - f(a)}{b - a}$

Here, $f(x) = \sqrt{x}$, $a = 25$, and $b = 289$.

Step 1: Compute the average rate of change

The average rate of change of $f(x)$ on $[25, 289]$ is: $\frac{f(289) - f(25)}{289 - 25}$ Substitute the values: $f(289) = \sqrt{289} = 17, \quad f(25) = \sqrt{25} = 5$ $\text{Average rate of change} = \frac{17 - 5}{289 - 25} = \frac{12}{264} = \frac{1}{22}$

Step 2: Find $f'(x)$

The derivative of $f(x) = \sqrt{x}$ is: $f'(x) = \frac{1}{2\sqrt{x}}$

Step 3: Set $f'(c) = \frac{1}{22}$

We need to solve: $\frac{1}{2\sqrt{c}} = \frac{1}{22}$ Multiply both sides by $2\sqrt{c}$: $1 = \frac{2\sqrt{c}}{22}$ Simplify: $\sqrt{c} = 11$ Square both sides: $c = 121$

Final Answer

The point $c$ that satisfies the conclusion of the MVT is: $c = 121$

Would you like further clarification or have any additional questions?

Related Questions:

1. What is the geometric interpretation of the Mean Value Theorem?
2. How would the conclusion change if the function was $y(x) = x^2$ on the same interval?
3. Why is differentiability required for applying the MVT?
4. What would happen if the interval was changed to $[25, 400]$?
5. Can you use the MVT to prove the inequality $\sqrt{x} \leq \sqrt{y}$ for $x \leq y$?

Tip:

Always check both continuity and differentiability conditions before applying the Mean Value Theorem.