Verify mean value theorem for the function `f(x)=sqrt(25-x^(2))` in `[1,5]`

Explanation

To verify the Mean Value Theorem (MVT) for the function \( f(x) = \sqrt{25 - x^2} \) on the interval \([1, 5]\), we will follow these steps:

### Step 1: Check Continuity
The function \( f(x) = \sqrt{25 - x^2} \) is defined when \( 25 - x^2 \geq 0 \). This implies:

\[ x^2 \leq 25 \implies -5 \leq x \leq 5 \]

Since the interval \([1, 5]\) is a subset of \([-5, 5]\), the function is continuous on \([1, 5]\).

### Step 2: Check Differentiability
Next, we need to check if \( f(x) \) is differentiable on the interval \((1, 5)\). We find the derivative using the chain rule:

\[ f'(x) = \frac{d}{dx} \left( \sqrt{25 - x^2} \right) = \frac{1}{2\sqrt{25 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{25 - x^2}} \]

The derivative \( f'(x) \) exists for all \( x \) in the interval \((1, 5)\) since \( 25 - x^2 > 0 \) in this interval. Therefore, \( f(x) \) is differentiable on \((1, 5)\).

### Step 3: Apply Mean Value Theorem
According to the Mean Value Theorem, there exists at least one point \( c \) in the interval \((1, 5)\) such that:

\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

where \( a = 1 \) and \( b = 5 \).

#### Step 3.1: Calculate \( f(a) \) and \( f(b) \)
\[ f(1) = \sqrt{25 - 1^2} = \sqrt{24} \]
\[ f(5) = \sqrt{25 - 5^2} = \sqrt{0} = 0 \]

#### Step 3.2: Compute the Right-Hand Side
\[ \frac{f(b) - f(a)}{b - a} = \frac{0 - \sqrt{24}}{5 - 1} = \frac{-\sqrt{24}}{4} = -\frac{\sqrt{24}}{4} = -\frac{\sqrt{4 \cdot 6}}{4} = -\frac{2\sqrt{6}}{4} = -\frac{\sqrt{6}}{2} \]

### Step 4: Set the Derivative Equal to the Right-Hand Side
Now, we set \( f'(c) \) equal to the computed value:

\[ f'(c) = \frac{-c}{\sqrt{25 - c^2}} = -\frac{\sqrt{6}}{2} \]

This leads to the equation:

\[ \frac{c}{\sqrt{25 - c^2}} = \frac{\sqrt{6}}{2} \]

### Step 5: Solve for \( c \)
Cross-multiplying gives:

\[ 2c = \sqrt{6} \sqrt{25 - c^2} \]

Squaring both sides:

\[ 4c^2 = 6(25 - c^2) \]
\[ 4c^2 = 150 - 6c^2 \]
\[ 10c^2 = 150 \]
\[ c^2 = 15 \]
\[ c = \sqrt{15} \quad (\text{since } c \text{ must be positive}) \]

### Conclusion
Since \( \sqrt{15} \) is approximately \( 3.87 \), which is in the interval \((1, 5)\), we have verified the Mean Value Theorem for the function \( f(x) = \sqrt{25 - x^2} \) on the interval \([1, 5]\).