Animated Solution for Mathematics - Differentiation: Let $f(x)$ be defined in the interval $[-2,2]$ such that $f(x)=\{\begin{array}{ll}-1,& -2\le x\le 0\\ x-1,& 0<x\le 2\end{array}$ and $g(x)=f(|x|)+|f(x)|$. Test the differentiability of $g(x)$ in $(-2,2)$.

A Journey into Differentiability

Welcome, future engineers. Today, we are going to dissect a problem that separates the casual students from the masters of calculus. We are looking at the function $g(x)=f(|x|)+|f(x)|$.

It looks innocent, doesn't it? But within this simple expression lies a trap that catches many students off guard. The key to mastering this is to stop seeing it as a single, monolithic function and start seeing it as a collection of behaviors.

We are going to break this down, interval by interval, until the function reveals its true nature.

The Deconstruction

First, let's talk about the absolute value. It is the most common source of "sharp corners" in calculus. When you see $|x|$, you should immediately think: "This is a switch." It switches behavior at $x=0$.

When you see $|f(x)|$, you should think: "This is another switch." It switches behavior whenever $f(x)$ crosses the $x$-axis. In our case, $f(x)=x-1$ for $x>0$, so it crosses at $x=1$.

So, we have two critical points: $x=0$ and $x=1$. These are the points where our function might "break" its smoothness. We will analyze the interval $[-2,0]$, then $(0,1]$, and finally $(1,2]$.

The Interval Analysis

Let's dive into the first interval, $x\in [-2,0]$. Here, $f(|x|)=f(-x)$. Since $-x$ is positive, we use the second branch of our definition:

Now, what about $|f(x)|$? In this interval, $f(x)=-1$, so $|f(x)|=|-1|=1$. Adding these together, we get:

This is a beautiful, linear segment. But wait! What happens at $x=0$? The derivative is $-1$.

Now, let's move to the second interval, $(0,1]$. Here, $f(|x|)=f(x)=x-1$. And $|f(x)|=|x-1|$.

Since $x\le 1$, $x-1$ is negative, so $|x-1|=-(x-1)=1-x$. Adding these together, we get:

This is a flat, horizontal line. The derivative is $0$. Do you see the problem? At $x=0$, the slope jumps from $-1$ to $0$. That is a sharp corner; it is not differentiable.

Finally, let's move to the last interval, $(1,2]$. Here, $f(|x|)=x-1$ and $|f(x)|=x-1$. Adding them gives:

The derivative is $2$. At $x=1$, the slope jumps from $0$ to $2$. Another sharp corner! Another point of non-differentiability.

The Conclusion

This is the beauty of calculus. We didn't need a complex formula; we just needed to understand the geometry of the function.

We have proven that $g(x)$ is continuous, but it has two "kinks" where the derivative fails to exist. Keep this mindset: when in doubt, break it down. Piecewise functions are just puzzles waiting to be solved.

You have successfully navigated the trap. Remember, in JEE Advanced, the ability to visualize the function's behavior is just as important as the algebraic manipulation. You have done well.