Animated Solution for Mathematics - Limits, Continuity and Differentiability: The function $f(x)=\{\begin{array}{ll}|x-3|,& x\ge 1\\ \frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4},& x<1\end{array}$ is

Imagine you are walking along a path. For the first part of your journey, you are following a smooth, gentle curve defined by a quadratic equation.

Suddenly, at a specific marker, the path changes character, becoming a sharp, V-shaped trail defined by a modulus function. This is exactly what we are dealing with in this problem.

Piecewise functions are like stories with two chapters; our job is to ensure the transition between these chapters is seamless.

The Meeting Point at $x=1$

Our first task is to see if the two paths actually meet at $x=1$. If they do not, there is a jump, and the function is discontinuous.

We calculate the Left-Hand Limit (LHL) using the quadratic part:

Substituting $x=1$, we get:

Now, we check the Right-Hand Limit (RHL) using the modulus part: $f(x)=|x-3|$. At $x=1$, this is $|1-3|=|-2|=2$.

Since the LHL, RHL, and the function value $f(1)$ are all equal to $2$, the paths meet perfectly. The function is continuous at $x=1$.

The Smoothness Test at $x=1$

Continuity is great, but is the transition smooth? This is where differentiability comes in.

We need to check if the slope of the quadratic curve matches the slope of the modulus line at the exact moment they meet. Differentiating the quadratic part, we get:

At $x=1$, the slope is:

Now, for the modulus part, since $x$ is near $1$ (and thus $x<3$), the expression $|x-3|$ behaves like $-(x-3)=3-x$. The derivative of $3-x$ is simply $-1$.

Because the slopes match (both are $-1$), the transition is perfectly smooth. The function is differentiable at $x=1$.

The Sharp Corner at $x=3$

Finally, we look at $x=3$. This is the heart of the modulus function.

We know that $|x-3|$ is continuous everywhere, so it is definitely continuous at $x=3$. However, differentiability is a different story.

If we approach $x=3$ from the left, the function is $3-x$, and the slope is $-1$. If we approach from the right, the function is $x-3$, and the slope is $1$.

Because $-1 eq 1$, there is a sharp 'kink' in the graph. This sharp corner means the function is not differentiable at $x=3$.

We have successfully navigated the function's behavior. We found that it is continuous at both $x=1$ and $x=3$, and differentiable at $x=1$.

However, the sharp corner at $x=3$ prevents it from being differentiable there. This problem teaches us that while continuity is about connection, differentiability is about the elegance of a smooth, unbroken slope.