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 need to follow these steps:

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

**Inequality:**
\[ 25 - x^2 \geq 0 \implies x^2 \leq 25 \implies -5 \leq x \leq 5 \]

Since the interval \([1, 5]\) lies within the domain of \( f(x) \), the function is continuous on \([1, 5]\).

### Step 2: Check Differentiability
Next, we need to find the derivative of \( f(x) \).

**Derivative Calculation:**
Using the chain rule,
\[ f'(x) = \frac{d}{dx} \left( \sqrt{25 - x^2} \right) = \frac{-x}{\sqrt{25 - x^2}} \]

This derivative exists for all \( x \) in the open interval \( (1, 5) \).

### Step 3: Apply Mean Value Theorem
According to the Mean Value Theorem, there exists at least one \( c \in (1, 5) \) such that:
\[ f'(c) = \frac{f(5) - f(1)}{5 - 1} \]

**Calculating \( f(5) \) and \( f(1) \):**
\[ f(5) = \sqrt{25 - 5^2} = \sqrt{0} = 0 \]
\[ f(1) = \sqrt{25 - 1^2} = \sqrt{24} = 2\sqrt{6} \]

**Substituting into MVT:**
\[ f'(c) = \frac{0 - 2\sqrt{6}}{4} = -\frac{\sqrt{6}}{2} \]

### Step 4: Set Up the Equation
Now, we set \( f'(c) \) equal to \(-\frac{\sqrt{6}}{2}\):
\[ -\frac{c}{\sqrt{25 - c^2}} = -\frac{\sqrt{6}}{2} \]

### Step 5: Solve for \( c \)
Cross-multiplying gives:
\[ 2c = \sqrt{6} \sqrt{25 - c^2} \]

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

### Step 6: Verify \( c \) is in the Interval
Now we check if \( c = \sqrt{15} \) lies within the interval \((1, 5)\):
\[ 1 < \sqrt{15} < 5 \]

Since \( \sqrt{15} \approx 3.87 \), it is indeed in the interval \((1, 5)\).

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

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