Animated Solution for Mathematics - Limits, Continuity and Differentiability: Determine the values of $x$ for which the following function fails to be continuous or differentiable: $f(x)=\{\begin{array}{ll}1-x,& x<1\\ (1-x)(2-x),& 1\le x\le 2\\ 3-x,& x>2\end{array}$ Justify your answer.

A Journey Through Piecewise Functions

Welcome, students. Today, we are not just solving a problem; we are performing an autopsy on a function. We are looking at a piecewise function, $f(x)$, which changes its identity as we move along the $x$-axis.

It is defined as:

In the world of JEE Advanced, piecewise functions are like bridges. We have three distinct segments, and our job is to determine if this bridge is safe to walk on—that is, is it continuous? And is it smooth—that is, is it differentiable?

The Visual Intuition

Before we touch a single variable, let us visualize. For $x<1$, we have a simple line, $1-x$. It is descending with a slope of $-1$.

Then, between $x=1$ and $x=2$, we have a quadratic, $(1-x)(2-x)$, which expands to $x^2-3x+2$. This is a parabola opening upwards.

Finally, for $x>2$, we return to a line, $3-x$. The critical moments—the 'seams' of our bridge—are at $x=1$ and $x=2$. This is where the function might break or kink.

The Investigation at $x=1$

Let us test the first seam. To be continuous, the function must approach the same value from both sides.

For the Left Hand Limit (LHL) at $x=1$, we use the first piece:

For the Right Hand Limit (RHL) at $x=1$, we use the second piece:

Since the LHL equals the RHL, the bridge is connected! Now, we check for smoothness by comparing derivatives.

The derivative of the left piece ($1-x$) is $-1$. The derivative of the right piece ($x^2-3x+2$) is $2x-3$.

At $x=1$, the derivative of the parabola is $2(1)-3=-1$. Because the slopes match, the transition is perfectly smooth. At $x=1$, our function is both continuous and differentiable.

The Moment of Truth at $x=2$

Now, we move to the second seam. This is where many students rush and stumble. Let us be precise.

For the LHL at $x=2$, we use the parabola:

The graph is approaching the $x$-axis from the left. For the RHL at $x=2$, we use the third piece:

Stop. Look at those numbers. The LHL is $0$, but the RHL is $1$. The graph is trying to be at two different heights at the same time. This is a jump discontinuity. The bridge has snapped.

Because the function is discontinuous at $x=2$, it is mathematically impossible for it to be differentiable there. Differentiability requires a smooth, unbroken path.

Since the path is broken, the derivative does not exist. We have successfully identified the failure point.

The function fails to be continuous and differentiable at $x=2$. Remember, in calculus as in life, continuity is the prerequisite for everything else. Keep practicing, stay curious, and never stop questioning the 'why' behind the math.