Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f(x)=\{\begin{array}{ll}0,& x<0\\ x^2,& x\ge 0\end{array}$ then for all $x$

Welcome, fellow explorers of the mathematical landscape. Today, we are going to dissect a function that seems simple on the surface but hides a profound lesson about the nature of calculus.

We are looking at the piecewise function:

At first glance, it looks like a simple combination of a flat line and a parabola. We must determine if they meet in a clean, seamless weld or a jagged edge.

The Glue Test

To understand if a function is continuous, we must ask: does the graph have any holes or jumps? We examine the transition point, $x=0$.

The Left Hand Limit (LHL) as $x\to 0^{-}$ is $0$, because the function is constantly $0$ on that side. The Right Hand Limit (RHL) as $x\to 0^{+}$ is:

Since the function value $f(0)=0^2=0$ also matches, we have $LHL=RHL=f(0)$. The graph is perfectly glued together; there is no break.

The Smoothness Test

Now, we move to the more rigorous test: differentiability. Does the function have a unique tangent at $x=0$? We calculate the derivative piecewise.

For $x<0$, $f^{\prime }(x)=\frac{d}{dx}(0)=0$. For $x>0$, $f^{\prime }(x)=\frac{d}{dx}(x^2)=2x$.

Now, we check the one-sided derivatives at $x=0$. The Left Hand Derivative (LHD) is $0$. The Right Hand Derivative (RHD) is $2(0)=0$.

Because $LHD=RHD=0$, the slopes match perfectly. Geometrically, the parabola flattens out exactly as it hits the x-axis, meeting the flat line with zero slope. It is differentiable.

The Derivative's Dilemma

But what about the derivative function itself, $f^{\prime }(x)$? We have:

Is this new function continuous? The LHL is $0$, and the RHL is ${\lim }_{x\to 0^{+}}2x=0$. Since $f^{\prime }(0)=0$, the derivative function is indeed continuous.

But is it differentiable? This is where the trap lies. To check if $f^{\prime }(x)$ is differentiable, we look at its derivative, $f^{\prime \prime }(x)$.

For $x<0$, $f^{\prime \prime }(x)=0$. For $x>0$, $f^{\prime \prime }(x)=2$. As we approach $x=0$, the LHD of $f^{\prime }(x)$ is $0$, but the RHD of $f^{\prime }(x)$ is $2$.

They are not equal! This means there is a sharp corner in the graph of $f^{\prime }(x)$ at the origin. Even though $f^{\prime }(x)$ is continuous, it is not differentiable.

We have journeyed through the layers of this function. We found that $f(x)$ is continuous and differentiable everywhere.

We found that $f^{\prime }(x)$ is continuous everywhere, but not differentiable at $x=0$. This problem teaches us that smoothness is a hierarchy: a function can be smooth enough to have a derivative, but its derivative might not be smooth enough to have a second derivative.

Keep exploring, keep questioning, and remember: every sharp corner in a graph is just a place where the math is telling you something new.