Animated Solution for Mathematics - Differentiation: Let $f:\mathbb{R}\to \mathbb{R}$ be defined as $f(x)=|x|+|x^2-1|$. The total number of points at which $f$ attains either a local maximum or a local minimum is

Welcome, fellow explorer of the mathematical landscape. Today, we are going to dissect a function that, at first glance, looks like a simple sum of absolute values:

But beneath this innocent exterior lies a fascinating structure of peaks and valleys. Our goal is to find every single point where this function decides to turn around—the local maxima and minima.

The Critical Boundaries

Before we touch any calculus, we must understand the terrain. Absolute value functions are like mountain ranges; they have sharp, jagged peaks and deep, narrow valleys.

These sharp turns occur exactly where the expressions inside the modulus bars equal zero. For $|x|$, that happens at $x=0$. For $|x^2-1|$, we solve $x^2-1=0$, which gives us $x=1$ and $x=-1$.

These three points—$x=-1,0,1$—are our signposts. They divide the entire real number line into four distinct zones. We cannot analyze the function as a whole; we must analyze it zone by zone.

The Four Zones

Let's walk through these intervals one by one:

1. The Rightmost Frontier ($x\ge 1$): Here, both $x$ and $x^2-1$ are non-negative. The modulus signs vanish, leaving us with:

This is a simple parabola opening upwards. Its derivative is $f^{\prime }(x)=2x+1$. For $x\ge 1$, this derivative is always positive, meaning the function is strictly increasing. No peaks or valleys here.

2. The Inner Valley ($0\le x\le 1$): Here, $x$ is positive, but $x^2-1$ is negative. We must flip the sign of the second term:

This is a downward-opening parabola. Its derivative is $f^{\prime }(x)=-2x+1$. Setting this to zero gives $x=\frac{1}{2}$. Since $\frac{1}{2}$ is inside our interval, we have found a stationary point! Because it is a downward parabola, this is a local maximum.

3. The Mirror Zone ($-1\le x\le 0$): Now we cross the y-axis. Both $x$ and $x^2-1$ are negative. The function becomes:

Another downward parabola! Its derivative is $f^{\prime }(x)=-2x-1$. Setting this to zero gives $x=-\frac{1}{2}$. This is another valid stationary point and another local maximum.

4. The Leftmost Frontier ($x\le -1$): Finally, $x$ is negative, but $x^2-1$ is positive again. The function is:

The derivative is $f^{\prime }(x)=2x-1$. For $x\le -1$, this is always negative. No stationary points here.

The Sharp Corners

We have found two local maxima at $x=\frac{1}{2}$ and $x=-\frac{1}{2}$. But we are not done. We must return to our critical boundaries: $x=-1,0,1$.

At these points, the function is not differentiable—it has sharp corners. Let's check the slopes around them:

- At $x=0$, the slope changes from $-1$ to $1$. A change from negative to positive indicates a local minimum. - At $x=1$, the slope changes from $-1$ to $3$. Again, negative to positive, indicating a local minimum. - At $x=-1$, the slope changes from $-3$ to $1$. Another local minimum.

By carefully partitioning the domain and checking both the smooth stationary points and the sharp corners, we have uncovered the full story of the function.

We have two local maxima (the smooth peaks of our parabolas) and three local minima (the sharp V-shaped dips). Adding these together, we find a total of $2+3=5$ points of local extrema.

You have successfully navigated the landscape of this function. Keep this analytical rigor in your toolkit—it is the key to mastering JEE Advanced calculus.