Number of Local Extrema for Sum of Absolute Values Function

To find the total number of points at which the function $f(x) = |x| + |x^2 - 1|$ attains a local maximum or a local minimum, we analyze the function piecewise.

Step 1: Symmetry of the Function

Note that for any $x \in \mathbb{R}$: $f(-x) = |-x| + |(-x)^2 - 1| = |x| + |x^2 - 1| = f(x)$ Thus, $f(x)$ is an even function, which means it is symmetric about the line $x = 0$.

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Step 2: Piecewise Analysis for $x \ge 0$

We divide the domain $x \ge 0$ into intervals based on the absolute value definitions:

1. Interval $0 \le x < 1$: In this interval, $|x| = x$ and $|x^2 - 1| = 1 - x^2$. $f(x) = x + (1 - x^2) = -x^2 + x + 1$ Taking the first derivative: $f'(x) = -2x + 1$ Setting $f'(x) = 0$ gives the critical point: $x = \frac{1}{2}$

   • For $x \in \left(0, \frac{1}{2}\right)$, $f'(x) > 0$, so $f(x)$ is strictly increasing.
   • For $x \in \left(\frac{1}{2}, 1\right)$, $f'(x) < 0$, so $f(x)$ is strictly decreasing.

2. Interval $x \ge 1$: In this interval, $|x| = x$ and $|x^2 - 1| = x^2 - 1$. $f(x) = x + (x^2 - 1) = x^2 + x - 1$ Taking the first derivative: $f'(x) = 2x + 1$ For $x > 1$, $f'(x) > 0$, so $f(x)$ is strictly increasing on $[1, \infty)$.

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Step 3: Identification of Local Extrema for $x \ge 0$

• At $x = 0$: Since $f(x)$ is strictly increasing on $\left(0, \frac{1}{2}\right)$ and symmetric about $x = 0$, it is strictly decreasing on $\left(-\frac{1}{2}, 0\right)$. Therefore, $f(x)$ attains a local minimum at $x = 0$. $f(0) = 1$

• At $x = \frac{1}{2}$: The derivative $f'(x)$ changes sign from positive to negative as $x$ increases through $\frac{1}{2}$. Therefore, $f(x)$ attains a local maximum at $x = \frac{1}{2}$. $f\left(\frac{1}{2}\right) = \frac{5}{4}$

• At $x = 1$: The function is strictly decreasing on $\left(\frac{1}{2}, 1\right)$ and strictly increasing on $(1, \infty)$. Therefore, $f(x)$ attains a local minimum at $x = 1$. $f(1) = 1$

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Step 4: Using Symmetry for $x < 0$

By the even symmetry of $f(x)$:

• At $x = -\frac{1}{2}$, $f(x)$ attains a local maximum.
• At $x = -1$, $f(x)$ attains a local minimum.

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Step 5: Total Count of Local Extrema

The points of local extrema are:

1. $x = -1$ (Local Minimum)
2. $x = -\frac{1}{2}$ (Local Maximum)
3. $x = 0$ (Local Minimum)
4. $x = \frac{1}{2}$ (Local Maximum)
5. $x = 1$ (Local Minimum)

Thus, the total number of points at which $f$ attains a local minimum or local maximum is 5.