Question

The set of all critical points of the function \( f(x) = |x^2 - 1| \) on \( [-2, 2] \) is ...........

• A. \( \{-1, 1\} \)
• B. \( \{-1, 0, 1\} \)
• C. \( \{1, 0\} \)
• D. \( \{0\} \)

Hint:

To find the critical points of piecewise functions involving absolute values, consider the derivative of each piece separately and check where the function changes behavior (where the absolute value expression is zero).

Answer 1

The Correct Option is B

The function is given by \( f(x) = |x^2 - 1| \). To find the critical points, we need to check where the derivative of \( f(x) \) does not exist or is zero.

First, rewrite the function without the absolute value for easier analysis: \[ f(x) = \begin{cases} x^2 - 1 & \text{if } x^2 \geq 1 \\ 1 - x^2 & \text{if } x^2 < 1 \end{cases} \]
Now, find the derivative for each piece:
- For \( x^2 - 1 \), the derivative is \( f'(x) = 2x \).
- For \( 1 - x^2 \), the derivative is \( f'(x) = -2x \).

Next, we find the points where the derivative is zero or undefined:
- \( f'(x) = 0 \) gives \( x = 0 \) for both cases.
- The function \( f(x) = |x^2 - 1| \) has nondifferentiable points where the expression inside the absolute value equals zero, i.e., at \( x = -1 \) and \( x = 1 \).

Thus, the critical points are \( x = -1 \), \( x = 0 \), and \( x = 1 \). Therefore, the set of all critical points on \( [-2, 2] \) is \( \{-1, 0, 1\} \).