Animated Solution for Mathematics - Functions: If $f(x)=\{\begin{array}{ll}2+2x,& -1\le x<0\\ 1-\frac{x}{3},& 0\le x\le 3\end{array};g(x)=\{\begin{array}{ll}-x,& -3\le x\le 0\\ x,& 0<x\le 1\end{array}$, then range of $(f\circ g(x))$ is

Welcome, future engineer! Today, we are going to demystify one of the most elegant concepts in calculus: the composite function.

When you see $f(g(x))$, I want you to stop thinking of it as a scary algebraic expression and start thinking of it as a machine. Imagine a two-stage factory where the first machine, $g(x)$, processes raw material $x$ into a product that is fed directly into the second machine, $f(x)$, to produce the final output.

Our goal is to find the range of this entire factory—the set of all possible final products.

Analyzing the Inner Machine

We begin with the inner function, $g(x)$, which is defined piecewise:

For the interval $x\in [-3,0]$, the function is $g(x)=-x$. This line starts at $(-3,3)$ and descends to the origin $(0,0)$, meaning the output values span from $0$ to $3$.

For the second interval, $x\in (0,1]$, the function is $g(x)=x$. This line rises from the origin to $(1,1)$, spanning output values from $0$ to $1$.

When we combine these two, we realize that the inner machine $g(x)$ is capable of producing any value in the interval $[0,3]$. This is the "fuel" that will be fed into our second machine.

The Bridge

This is where many students stumble. Instead of solving for $x$ in the composite function directly, use the range of $g(x)$ as the domain of $f(x)$.

We have established that $g(x)\in [0,3]$. Now, we look at the outer function $f(x)$:

Since our input (the output of $g(x)$) is restricted to $[0,3]$, we only care about the second branch of $f(x)$. The first branch, defined for $x\in [-1,0)$, is completely irrelevant because our machine $g(x)$ never outputs a negative number.

It is like having a machine that only accepts inputs of a certain voltage; if the first stage never provides that voltage, the second stage never activates. Thus, we focus entirely on $f(x)=1-\frac{x}{3}$ for $x\in [0,3]$.

The Final Calculation

Now, we substitute our range $[0,3]$ into the active branch of $f(x)$. We are looking for the range of $f(g(x))=1-\frac{g(x)}{3}$.

Since $g(x)$ varies from $0$ to $3$, let's test the boundaries: When $g(x)=0$, $f(0)=1-\frac{0}{3}=1$. When $g(x)=3$, $f(3)=1-\frac{3}{3}=0$.

Because $f(x)$ is a linear function, it is continuous and monotonic. It will smoothly transition between these two values.

Therefore, the output of the entire composite system covers every value from $0$ to $1$. The final range is $[0,1]$.

See how simple it becomes when you break it down? You didn't need complex algebra; you needed a clear mental map of the flow of values.

You analyzed the inner machine, identified the relevant branch of the outer machine, and mapped the transformation. This is the heart of JEE Advanced problem-solving: not brute force, but clear, logical visualization.

Keep practicing this "machine" mindset, and you will find that even the most intimidating functions become easy to conquer. You've got this!