Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f(x)=\{\begin{array}{l}1+x,0\le x\le 2\\ 3-x,2<x\le 3\end{array}$. Determine the form of $g(x)=f(f(x))$ and hence find the points of discontinuity of $g$, if any.

Welcome, fellow traveler on the journey of JEE Advanced mathematics. Today, we are going to dissect a problem that often trips up even the most diligent students: the composition of piecewise functions.

It is not just about plugging one equation into another; it is about understanding the 'machine' that is a function.

Imagine $f(x)$ as a machine. You feed it a number $x$, and it spits out a value $f(x)$.

Now, consider $g(x)=f(f(x))$. This is a two-stage machine where you feed $x$ into the first machine, and the output is immediately fed into the second machine.

The critical realization here is that the output of the first machine must be a valid input for the second. If the first machine produces a value that the second machine cannot process, the process breaks down. This is why we must be meticulous about the range of the inner function.

Our function is defined as:

Notice the boundary at $x=2$. This is the 'switch' in our machine.

When we look at $g(x)=f(f(x))$, the rule changes whenever the inner $f(x)$ crosses this boundary. To find these points, we solve $f(x)=2$.

For the first branch, $1+x=2$ gives $x=1$. For the second branch, $3-x=2$ gives $x=1$ (which is outside the domain) and $x=2$ is the boundary. Thus, our critical points are $x=1$ and $x=2$.

Let us break this down into intervals.

For $x\in [0,1]$, $f(x)=1+x$, which ranges from $1$ to $2$. Since this output is within the $[0,2]$ range of the outer function, we use the first branch:

For $x\in (1,2]$, $f(x)=1+x$, which ranges from $2$ to $3$. This output is now in the $(2,3]$ range of the outer function, so we must use the second branch:

Finally, for $x\in (2,3]$, $f(x)=3-x$, which ranges from $0$ to $1$. This is back in the first branch range:

Continuity

Now, we check for continuity. At $x=1$, the left-hand limit is $2+1=3$, and the right-hand limit is $2-1=1$.

Since $3 eq 1$, there is a jump discontinuity.

At $x=2$, the left-hand limit is $2-2=0$, and the right-hand limit is $4-2=2$. Since $0 eq 2$, there is another jump discontinuity.

Mathematics is often about finding the hidden structure in chaos. By tracking the domain and range, we have tamed this composite function. Keep practicing, and soon, you will see these patterns instantly!