The value of x in the interval [4,9] at which the function f(x) = `sqrtx` satisfies the mean value theoram is

Options

A. `(13)/(4)`
B. `(17)/(4)`
C. `(21)/(4)`
D. `(25)/(4)`

Answer

To solve the problem, we will follow the steps outlined in the video transcript and apply the Mean Value Theorem (MVT) to the function \( f(x) = \sqrt{x} \) over the interval \([4, 9]\).

### Step-by-Step Solution:

1. **Identify the Function and Interval**:
- The function is \( f(x) = \sqrt{x} \).
- The closed interval is \([4, 9]\).

2. **Check Continuity and Differentiability**:
- The function \( f(x) = \sqrt{x} \) is continuous on the interval \([4, 9]\) because square root functions are continuous for \( x \geq 0 \).
- The function is also differentiable on the open interval \( (4, 9) \) since the derivative exists for all \( x > 0 \).

3. **Apply the Mean Value Theorem**:
- According to the Mean Value Theorem, there exists at least one point \( c \) in the open interval \( (4, 9) \) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
- Here, \( a = 4 \) and \( b = 9 \).

4. **Calculate \( f(a) \) and \( f(b) \)**:
- \( f(4) = \sqrt{4} = 2 \)
- \( f(9) = \sqrt{9} = 3 \)

5. **Calculate the Right-Hand Side of the MVT**:
- Substitute \( f(a) \) and \( f(b) \) into the MVT formula:
\[ f'(c) = \frac{f(9) - f(4)}{9 - 4} = \frac{3 - 2}{5} = \frac{1}{5} \]

6. **Find the Derivative of \( f(x) \)**:
- The derivative of \( f(x) = \sqrt{x} \) is:
\[ f'(x) = \frac{1}{2\sqrt{x}} \]

7. **Set the Derivative Equal to the MVT Result**:
- Set \( f'(c) \) equal to \( \frac{1}{5} \):
\[ \frac{1}{2\sqrt{c}} = \frac{1}{5} \]

8. **Solve for \( c \)**:
- Cross-multiply to solve for \( c \):
\[ 5 = 2\sqrt{c} \]
- Divide both sides by 2:
\[ \sqrt{c} = \frac{5}{2} \]
- Square both sides:
\[ c = \left(\frac{5}{2}\right)^2 = \frac{25}{4} \]

9. **Conclusion**:
- The value of \( c \) that satisfies the Mean Value Theorem in the interval \([4, 9]\) is:
\[ c = \frac{25}{4} = 6.25 \]

### Final Answer:
The value of \( x \) in the interval \([4, 9]\) at which the function \( f(x) = \sqrt{x} \) satisfies the Mean Value Theorem is \( c = 6.25 \).

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