To solve the problem, we need to analyze the function \( f(x) = |x^2 - 1| \) over the interval \([-2, 2]\) and verify the statements provided.
### Step 1: Analyze the function \( f(x) \)
The function is defined as:
\[
f(x) = |x^2 - 1|
\]
We need to determine where \( x^2 - 1 \) is positive and where it is negative.
### Step 2: Find the points where \( x^2 - 1 = 0 \)
Setting the expression inside the absolute value to zero:
\[
x^2 - 1 = 0 \implies x^2 = 1 \implies x = \pm 1
\]
Thus, the critical points are \( x = -1 \) and \( x = 1 \).
### Step 3: Determine the intervals
We can analyze the sign of \( x^2 - 1 \) in the intervals:
1. \( (-\infty, -1) \)
2. \( (-1, 1) \)
3. \( (1, \infty) \)
- For \( x < -1 \) (e.g., \( x = -2 \)):
\[
x^2 - 1 = 4 - 1 = 3 \quad \text{(positive)}
\]
So, \( f(x) = x^2 - 1 \).
- For \( -1 < x < 1 \) (e.g., \( x = 0 \)):
\[
x^2 - 1 = 0 - 1 = -1 \quad \text{(negative)}
\]
So, \( f(x) = -(x^2 - 1) = 1 - x^2 \).
- For \( x > 1 \) (e.g., \( x = 2 \)):
\[
x^2 - 1 = 4 - 1 = 3 \quad \text{(positive)}
\]
So, \( f(x) = x^2 - 1 \).
### Step 4: Define \( f(x) \) piecewise
Thus, we can write \( f(x) \) as:
\[
f(x) =
\begin{cases}
x^2 - 1 & \text{if } x \leq -1 \\
1 - x^2 & \text{if } -1 < x < 1 \\
x^2 - 1 & \text{if } x \geq 1
\end{cases}
\]
### Step 5: Calculate the derivative \( f'(x) \)
Now, we differentiate \( f(x) \) in each interval:
1. For \( x \leq -1 \):
\[
f'(x) = 2x
\]
2. For \( -1 < x < 1 \):
\[
f'(x) = -2x
\]
3. For \( x \geq 1 \):
\[
f'(x) = 2x
\]
### Step 6: Check differentiability at critical points
At \( x = -1 \) and \( x = 1 \), we need to check the derivatives:
- \( f'(-1) \) from the left: \( 2(-1) = -2 \)
- \( f'(-1) \) from the right: \( -2(-1) = 2 \)
Since the derivatives do not match, \( f(x) \) is not differentiable at \( x = -1 \).
Similarly, check at \( x = 1 \):
- \( f'(1) \) from the left: \( -2(1) = -2 \)
- \( f'(1) \) from the right: \( 2(1) = 2 \)
Again, the derivatives do not match, so \( f(x) \) is not differentiable at \( x = 1 \).
### Step 7: Apply Rolle's Theorem
Since \( f(-2) = f(2) \) and \( f(x) \) is continuous on \([-2, 2]\) and differentiable on \((-2, -1)\), \((-1, 1)\), and \((1, 2)\), by Rolle's Theorem, there exists at least one \( c \) in \((-2, 2)\) such that \( f'(c) = 0 \).
### Step 8: Evaluate \( f'(0) \)
Since \( 0 \) is in the interval \((-1, 1)\):
\[
f'(0) = -2(0) = 0
\]
### Conclusion
Both statements are true:
- Statement 1 is true since there exists at least one \( c \) such that \( f'(c) = 0 \).
- Statement 2 is true since \( f'(0) = 0 \).
