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 defined as the absolute value of \( x^2 - 1 \). To analyze the function, we consider two cases for \( x^2 - 1 \):

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

This piecewise function describes a parabola that is reflected along the x-axis when \( |x| < 1 \) and a parabola opening upwards for \( |x| \geq 1 \).

Step 2: Local minima occur where the function reaches its lowest value. First, notice that \( f(x) = 0 \) 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 transitions from decreasing to increasing, indicating that these are points of local minima.

Step 3: Local maxima occur where the function reaches its highest value within a given interval. Notice that the function reaches a local maximum at \( x = 0 \), because:

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

The function \( f(x) \) decreases on the interval \( (-1, 1) \) and then increases after \( x = \pm 1 \), so \( x = 0 \) is a local maximum.

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

Answer 2

We are given the function $f(x) = \cos x - 1 + \frac{x^2}{2}$ for $x \in \mathbb{R}$. We need to determine if the function is increasing, decreasing, neither increasing nor decreasing, or constant for $x > 0$ 

Step 1: Find the first derivative $f'(x)$

$$ f(x) = \cos x - 1 + \frac{x^2}{2} $$

$$ f'(x) = -\sin x + x = x - \sin x $$

Step 2: Analyze the sign of $f'(x)$

Consider the function $g(x) = f'(x) = x - \sin x$. Let's find its derivative:

$$ g'(x) = 1 - \cos x $$

We know that $-1 \le \cos x \le 1$ for all $x \in \mathbb{R}$. Therefore, $1 - \cos x \ge 0$ for all $x \in \mathbb{R}$. This means $g(x) = f'(x)$ is a non-decreasing function.

Step 3: Evaluate $f'(0)$

$$ f'(0) = 0 - \sin 0 = 0 $$

Step 4: Determine the sign of $f'(x)$ for $x > 0$

Since $f'(x)$ is non-decreasing and $f'(0) = 0$, for $x > 0$, we have $f'(x) \ge f'(0) = 0$. In fact, for $x > 0$, $x > \sin x$, so $f'(x) = x - \sin x > 0$. This implies $f(x)$ is increasing for $x > 0$.

Step 5: Determine the sign of $f'(x)$ for $x < 0$

For $x < 0$, let $x = -y$ where $y > 0$.

$$ f'(-y) = -y - \sin(-y) = -y + \sin y = -(y - \sin y) $$

Since $y > 0$, $y - \sin y > 0$, so $f'(-y) < 0$. This implies $f(x)$ is decreasing for $x < 0$.

Step 6: Conclusion about the monotonicity of $f(x)$

The function $f(x)$ is decreasing for $x < 0$ and increasing for $x > 0$. Therefore, over its entire domain $\mathbb{R}$, the function is neither increasing nor decreasing.

Final Answer: (C) neither increasing nor decreasing