Rolle's theorem is true for the function `f(x)=x^2-4` in the interval

Options

A. `[-2,0]`
B. `[-2,2]`
C. `[0,1/2]`
D. `[0,2]`

Answer

To determine if Rolle's theorem is applicable to the function \( f(x) = x^2 - 4 \) in a given interval, we need to verify the conditions of Rolle's theorem. Let's go through the steps systematically.

### Step 1: Identify the function and the interval
We have the function:
\[ f(x) = x^2 - 4 \]
We need to check this function in various intervals to see if it satisfies the conditions of Rolle's theorem.

### Step 2: Check for continuity
Rolle's theorem requires that the function is continuous on the closed interval \([a, b]\). The function \( f(x) = x^2 - 4 \) is a polynomial, and polynomials are continuous everywhere. Therefore, \( f(x) \) is continuous on any interval \([a, b]\).

**Hint:** Check if the function is a polynomial; if it is, it is continuous everywhere.

### Step 3: Check for differentiability
Next, we need to check if the function is differentiable on the open interval \((a, b)\). Since \( f(x) \) is a polynomial, it is also differentiable everywhere. Thus, \( f(x) \) is differentiable on any interval \((a, b)\).

**Hint:** Polynomials are differentiable everywhere, which means you can assume differentiability in any interval.

### Step 4: Check the values at the endpoints
Now we need to find the values of \( f(a) \) and \( f(b) \) and check if they are equal:
\[ f(a) = a^2 - 4 \]
\[ f(b) = b^2 - 4 \]
For Rolle's theorem to hold, we need:
\[ f(a) = f(b) \implies a^2 - 4 = b^2 - 4 \implies a^2 = b^2 \]
This implies that \( a = b \) or \( a = -b \).

**Hint:** Set the values of the function at the endpoints equal to each other to check if they satisfy the condition.

### Step 5: Find the critical point
Next, we find the derivative of the function:
\[ f'(x) = 2x \]
To find the critical points, we set the derivative equal to zero:
\[ 2x = 0 \implies x = 0 \]
This critical point \( c = 0 \) must lie within the interval \((a, b)\).

**Hint:** Differentiate the function and set the derivative to zero to find critical points.

### Step 6: Choose an interval and verify
Now, we can check specific intervals to see if they satisfy all conditions. We will check the following intervals:
1. \([-2, 2]\)
2. \([-2, 0]\)
3. \([0, \frac{1}{2}]\)
4. \([0, 2]\)

**Interval 1: \([-2, 2]\)**
- \( a = -2, b = 2 \)
- \( f(-2) = (-2)^2 - 4 = 0 \)
- \( f(2) = (2)^2 - 4 = 0 \)
- \( f(-2) = f(2) \) (Condition satisfied)
- \( c = 0 \) is in \((-2, 2)\) (Condition satisfied)

**Interval 2: \([-2, 0]\)**
- \( a = -2, b = 0 \)
- \( f(-2) = 0 \)
- \( f(0) = -4 \)
- \( f(-2) \neq f(0) \) (Condition not satisfied)

**Interval 3: \([0, \frac{1}{2}]\)**
- \( a = 0, b = \frac{1}{2} \)
- \( f(0) = -4 \)
- \( f(\frac{1}{2}) = -\frac{15}{4} \)
- \( f(0) \neq f(\frac{1}{2}) \) (Condition not satisfied)

**Interval 4: \([0, 2]\)**
- \( a = 0, b = 2 \)
- \( f(0) = -4 \)
- \( f(2) = 0 \)
- \( f(0) \neq f(2) \) (Condition not satisfied)

### Conclusion
The only interval that satisfies all conditions of Rolle's theorem is \([-2, 2]\).

**Final Answer:** Rolle's theorem is true for the function \( f(x) = x^2 - 4 \) in the interval \([-2, 2]\).