Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f(x)=\{\begin{array}{ll}\frac{x^2}{2},& 0\le x<1\\ 2x^2-3x+\frac{3}{2},& 1\le x\le 2\end{array}$. Discuss the continuity of $f,f^{\prime }$ and $f^{\prime \prime }$ on $[0,2]$.

Imagine you are walking along a path defined by a mathematical function. As you traverse the interval $[0,2]$, you encounter a junction at $x=1$.

To your left, the path is governed by the simple parabola $f(x)=\frac{x^2}{2}$. To your right, the landscape shifts to a more complex quadratic, $f(x)=2x^2-3x+\frac{3}{2}$.

The question before us is simple yet profound: Is this path seamless, or are there hidden fractures in its construction?

Before we worry about how the path bends, we must ensure the path actually exists. Continuity at $x=1$ requires that the path from the left meets the path from the right at the exact same height.

We calculate the Left Hand Limit (LHL) as we approach $x=1$ from the left:

Now, we look at the Right Hand Limit (RHL) from the other side:

Since the LHL equals the RHL, and both equal the function value $f(1)=\frac{1}{2}$, we can breathe a sigh of relief. The path is continuous. There is no 'jump' or 'hole' in our journey.

Now that we know the path is connected, we must ask: is it smooth? A sharp corner would imply that the derivative is discontinuous.

To find out, we differentiate our piecewise function. For $0\le x<1$, the slope is $f^{\prime }(x)=x$. For $1<x\le 2$, the slope is $f^{\prime }(x)=4x-3$.

Let us test the junction again:

Because the Left Hand Derivative (LHD) matches the Right Hand Derivative (RHD), the transition is perfectly smooth. There is no sudden change in direction. You could roll a marble across this junction without it hitting a bump.

Here is where the story takes a fascinating turn. We have established that the path is continuous and smooth, but what about the 'bend' of the curve?

This is the realm of the second derivative, $f^{\prime \prime }(x)$. We differentiate our slopes once more:

For $0\le x<1$, $f^{\prime \prime }(x)=\frac{d}{dx}(x)=1$.

For $1<x\le 2$, $f^{\prime \prime }(x)=\frac{d}{dx}(4x-3)=4$.

As we approach $x=1$ from the left, the curvature is $1$. As we approach from the right, the curvature jumps to $4$. Even though the path is smooth, the way it curves changes instantaneously.

This is a discontinuity in the second derivative.

This problem teaches us a vital lesson in the hierarchy of calculus. Continuity of $f(x)$ and $f^{\prime }(x)$ does not guarantee the continuity of $f^{\prime \prime }(x)$.

We have successfully navigated the landscape, proving that while our path is connected and smooth, its underlying geometry undergoes a sudden, sharp change in concavity at $x=1$.

Keep this in mind as you tackle more complex problems—sometimes, the most interesting behavior happens in the derivatives we often overlook!