Question

Let \(f : \mathbb{R} \to \mathbb{R}\) be given by \(f(x) = |x^2 - 1|\), then:

• A. \(f\) has a local minima at \(x = 1\) but no local maxima
• B. \(f\) has a local maxima at \(x = 0\), but no local minima
• C. \(f\) has a local minima at \(x = \pm 1\) and a local maxima at \(x = 0\)
• D. \(f\) has neither any local maxima nor any local minima

Hint:

For absolute value functions, break the function into cases depending on the sign of the expression inside the absolute value. This will help you analyze the function’s behavior more easily and identify points of local maxima and minima.

Answer 1

The Correct Option is C

Step 1: The function \( f(x) = |x^2 - 1| \) is the absolute value of \( x^2 - 1 \). To analyze it, consider two cases:

\[ f(x) = \begin{cases} x^2 - 1 & \text{if } x^2 \geq 1, \\ 1 - x^2 & \text{if } x^2 < 1. \end{cases} \]

This piecewise function represents a parabola reflected across the x-axis when \( |x| < 1 \) and an upward-opening parabola when \( |x| \geq 1 \).

Step 2: Local minima are the function's lowest points. \( f(x) = 0 \) when \( x = \pm 1 \) because:

\[ f(x) = |x^2 - 1| = 0 \quad \text{when } x^2 - 1 = 0 \quad \implies \quad x = \pm 1. \]

At \( x = 1 \) and \( x = -1 \), the function changes from decreasing to increasing, thus representing local minima.

Step 3: Local maxima are the function's highest points within an interval. The function reaches a local maximum at \( x = 0 \):

\[ f(0) = |0^2 - 1| = | -1| = 1. \]

The function \( f(x) \) decreases on \( (-1, 1) \) and increases after \( x = \pm 1 \), making \( x = 0 \) a local maximum.

Step 4: Consequently, \( f(x) \) has local minima at \( x = \pm 1 \) and a local maximum at \( x = 0 \).