Animated Solution for Mathematics - Functions: Let $f$ and $g$ be two functions defined by $f(x)=\{\begin{array}{ll}x+1,& x<0\\ |x-1|,& x\ge 0\end{array}$ and $g(x)=\{\begin{array}{ll}x+1,& x<0\\ 1,& x\ge 0\end{array}$. Then $(g\circ f)(x)$ is

A Mathematical Journey

Imagine you are an architect designing a complex machine. You have two black boxes, $f(x)$ and $g(x)$. Your goal is to feed the output of the first machine into the second.

This is the essence of function composition: $(g\circ f)(x)=g(f(x))$. In the JEE Advanced arena, this is not just about plugging one equation into another; it is about understanding the 'hand-off' between two different mathematical behaviors.

Mapping the Terrain

We start with our two definitions:

To find $(g\circ f)(x)$, we must look at the outer function $g$. It changes its personality at $u=0$.

Therefore, the composition $g(f(x))$ will change its personality whenever $f(x)=0$. This is the hidden trap! Many students look at $x=0$ and stop there, but we must solve $f(x)=0$ to find the true transition points for the composite function.

The Search for Transition Points

Let's investigate where $f(x)<0$. For $x\ge 0$, $f(x)=|x-1|$, which is always $\ge 0$. So, no negative values exist in this domain.

For $x<0$, $f(x)=x+1$. Setting $x+1<0$ gives us $x<-1$.

This is our first major discovery: for all $x<-1$, the inner function $f(x)$ is negative, forcing the outer function $g$ to use its first branch, $g(u)=u+1$. Substituting $f(x)$ into this, we get $(g\circ f)(x)=(x+1)+1=x+2$.

The Horizontal Plateau

Now, what happens when $x\ge -1$? In this region, $f(x)\ge 0$.

Looking at our definition of $g(u)$, whenever the input is $\ge 0$, the output is simply $1$. Therefore, for the entire interval $x\ge -1$, our composite function collapses into a constant: $(g\circ f)(x)=1$.

We have successfully constructed our composite function:

Continuity and the Sharp Corner

Now, let's test the integrity of our creation at the junction $x=-1$.

To check continuity, we calculate the limits. The Left Hand Limit (LHL) as $x\to -1^{-}$ is ${\lim }_{x\to -1^{-}}(x+2)=1$. The Right Hand Limit (RHL) as $x\to -1^{+}$ is ${\lim }_{x\to -1^{+}}(1)=1$.

Since the LHL equals the RHL and the function value $f(-1)=1$, the function is perfectly continuous. There is no jump, no hole, and no break.

However, differentiability is a different beast. It asks: 'How smooth is the transition?'

The Left Hand Derivative (LHD) is the derivative of $x+2$, which is $1$. The Right Hand Derivative (RHD) is the derivative of $1$, which is $0$.

Because $1 eq 0$, the function has a sharp corner at $x=-1$. It is continuous, but it refuses to be smooth. It is like a road that turns abruptly—you can drive on it, but you cannot turn the steering wheel smoothly at that exact point.

The Beauty of the Result

We have navigated the logic, identified the transition point, and verified the behavior. The function $(g\circ f)(x)$ is continuous everywhere, but it fails to be differentiable at exactly one point: $x=-1$.

This problem teaches us that composition is not just algebra; it is a study of how boundaries interact. Keep this rigor in your toolkit, and you will find that even the most complex functions start to reveal their secrets.