Verify the truth of Rolle's Theorem for the following functions
`f(x)=x^(2)+2,a=-2,andb=2`

Answer

To verify the truth of Rolle's Theorem for the function \( f(x) = x^2 + 2 \) on the interval \( [a, b] = [-2, 2] \), we need to check the following conditions:

1. The function \( f(x) \) must be continuous on the closed interval \([a, b]\).
2. The function \( f(x) \) must be differentiable on the open interval \((a, b)\).
3. The values of the function at the endpoints must be equal, i.e., \( f(a) = f(b) \).
4. There exists at least one \( c \) in the open interval \((a, b)\) such that \( f'(c) = 0 \).

### Step 1: Check Continuity
The function \( f(x) = x^2 + 2 \) is a polynomial function. All polynomial functions are continuous everywhere. Therefore, \( f(x) \) is continuous on the closed interval \([-2, 2]\).

### Step 2: Check Differentiability
Similarly, since \( f(x) \) is a polynomial, it is also differentiable everywhere. Hence, \( f(x) \) is differentiable on the open interval \((-2, 2)\).

### Step 3: Check Endpoint Values
Now we calculate \( f(a) \) and \( f(b) \):
- \( f(-2) = (-2)^2 + 2 = 4 + 2 = 6 \)
- \( f(2) = (2)^2 + 2 = 4 + 2 = 6 \)

Since \( f(-2) = f(2) = 6 \), the third condition is satisfied.

### Step 4: Find \( c \) such that \( f'(c) = 0 \)
Next, we find the derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(x^2 + 2) = 2x \]
To find \( c \) such that \( f'(c) = 0 \):
\[ 2c = 0 \implies c = 0 \]
Since \( c = 0 \) lies in the open interval \((-2, 2)\), the fourth condition is satisfied.

### Conclusion
All conditions of Rolle's Theorem are satisfied for the function \( f(x) = x^2 + 2 \) on the interval \([-2, 2]\). Therefore, we can conclude that Rolle's Theorem holds true for this function.

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