Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f:R\to R$ be differentiable at $c\in R$ and $f(c)=0$. If $g(x)=|f(x)|$, then at $x=c$, $g$ is :

Imagine you are standing on a graph of a function $f(x)$. You are walking along the curve, and at the point $x=c$, you cross the x-axis. The value of the function is zero, $f(c)=0$.

Now, imagine a mirror placed along the x-axis. When we define a new function $g(x)=|f(x)|$, we are essentially taking everything below the x-axis and reflecting it upwards. This is the core of our problem.

In the world of calculus, we are obsessed with smoothness. We want our functions to be differentiable, which means they must have a unique, well-defined tangent line at every point. But reflection is a violent operation; it often takes a smooth, gentle curve and snaps it into a sharp, jagged corner.

Our mission today is to find the exact condition under which this reflection remains smooth.

Our Mathematical Microscope

To understand if $g(x)$ is differentiable at $x=c$, we cannot rely on intuition alone. We must use the most powerful tool in our arsenal: the first principle of derivatives. We define the derivative of $g$ at $c$ as the limit:

Since we know $f(c)=0$, it follows that $g(c)=|f(c)|=0$. The expression simplifies beautifully to:

This is the moment of truth. We need to know how $f(c+h)$ behaves as $h$ gets infinitesimally small. Because $f$ is differentiable at $c$, we can use a linear approximation.

Near $c$, the function $f$ behaves like a straight line with slope $f^{\prime }(c)$. Mathematically, we write this as $f(c+h)\approx f(c)+h{f}^{\prime }(c)$. Since $f(c)=0$, this becomes $f(c+h)\approx h{f}^{\prime }(c)$.

Now, substitute this approximation into our limit:

Here lies the heart of the problem. The term ${\lim }_{h\to 0}\frac{|h|}{h}$ is the gatekeeper of differentiability. Let us look at it from both sides.

If we approach from the right ($h>0$), then $|h|=h$, and the limit is $1$. If we approach from the left ($h<0$), then $|h|=-h$, and the limit is $-1$. For the derivative to exist, the left-hand derivative (LHD) must equal the right-hand derivative (RHD).

This means:

This equation implies $|f^{\prime }(c)|=-|f^{\prime }(c)|$. The only way a number can be equal to its own negative is if that number is zero. Therefore, $|f^{\prime }(c)|=0$, which forces $f^{\prime }(c)=0$.

What does $f^{\prime }(c)=0$ actually mean? It means that at the point where the function crosses the x-axis, the tangent line is horizontal.

Think of a parabola $f(x)=x^2$ at $x=0$. It touches the x-axis and turns around. When you take the absolute value, it remains $x^2$, which is perfectly smooth.

There is no sharp corner because the function was already 'flat' at the point of reflection. If the slope were non-zero, the reflection would create a 'V' shape, a cusp where the derivative is undefined.

By ensuring $f^{\prime }(c)=0$, we ensure that the transition is seamless. We have successfully navigated the trap of the absolute value, proving that smoothness is not just a visual quality, but a rigorous algebraic requirement.