In the Mean value theorem
`f(b)-f(a)=(b-a)f'(c),` if a=4, b=9
and f(x)=`sqrtx`, then the value of c is

Options

A. `8.00`
B. 5.25
C. `4.00`
D. 6.25

Answer

To find the value of \( c \) using the Mean Value Theorem for the function \( f(x) = \sqrt{x} \) over the interval \([4, 9]\), we will follow these steps:

### Step 1: Identify \( a \) and \( b \)
Given:
- \( a = 4 \)
- \( b = 9 \)

### Step 2: Calculate \( f(a) \) and \( f(b) \)
We need to find:
- \( f(4) = \sqrt{4} = 2 \)
- \( f(9) = \sqrt{9} = 3 \)

### Step 3: Apply the Mean Value Theorem
According to the Mean Value Theorem:
\[ f(b) - f(a) = (b - a) f'(c) \]
Substituting the values we have:
\[ 3 - 2 = (9 - 4) f'(c) \]
This simplifies to:
\[ 1 = 5 f'(c) \]

### Step 4: Find \( f'(x) \)
To find \( f'(c) \), we first need to differentiate \( f(x) \):
\[ f(x) = \sqrt{x} = x^{1/2} \]
Using the power rule:
\[ f'(x) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}} \]

### Step 5: Substitute \( f'(c) \) into the equation
Now we substitute \( f'(c) \) into our equation:
\[ 1 = 5 \cdot \frac{1}{2\sqrt{c}} \]
This simplifies to:
\[ 1 = \frac{5}{2\sqrt{c}} \]

### Step 6: Solve for \( \sqrt{c} \)
Rearranging gives:
\[ 2\sqrt{c} = 5 \]
Dividing both sides by 2:
\[ \sqrt{c} = \frac{5}{2} \]

### Step 7: Square both sides to find \( c \)
Now we square both sides:
\[ c = \left(\frac{5}{2}\right)^2 = \frac{25}{4} = 6.25 \]

### Conclusion
Thus, the value of \( c \) is:
\[ \boxed{6.25} \]