Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $f(x)=\{\begin{array}{ll}{\int }_0^x(5+|1-t|)dt,& x>2\\ 5x+1,& x\le 2\end{array}$, then

Navigating Piecewise Continuity

Welcome, fellow explorers of the mathematical universe. Today, we are going to dissect a problem that perfectly encapsulates the beauty and the traps of calculus.

We are looking at a piecewise function, $f(x)$, which behaves like a simple linear path for a while and then transforms into something more complex—an integral. Our goal is to determine the nature of this function at the junction point, $x=2$.

Is it a smooth transition, or is there a hidden sharp corner? Let us embark on this journey together.

The Linear Foundation

Imagine you are walking along a path defined by $f(x)=5x+1$ for all values of $x\le 2$. This is a straightforward, predictable linear journey.

At the exact moment you reach $x=2$, your position is:

The slope of this path is constant; it is the derivative of $5x+1$, which is simply $5$. This is our left-hand derivative, $f^{\prime }(2^{-})=5$. It is a steady, reliable slope.

The Integral Enigma

For $x>2$, our function is defined by an integral:

This looks intimidating, but let us break it down. The modulus function $|1-t|$ is the gatekeeper here, as it changes its behavior at $t=1$.

Because our upper limit $x$ is greater than $2$, our integral spans across the point $t=1$. To solve this, we must split the integral into two distinct regions: the region where $t\le 1$ and the region where $t>1$.

We write it as:

In the first region, $t\le 1$, so $|1-t|=1-t$. The integrand becomes $5+1-t=6-t$.

In the second region, $t>1$, so $|1-t|=t-1$. The integrand becomes $5+t-1=4+t$. Now, the problem is no longer a scary integral; it is just two simple polynomial integrations.

The Calculation

Let us evaluate the first part:

Now, the second part:

Adding these together:

Simplifying the constants, $\frac{11}{2}-\frac{9}{2}=\frac{2}{2}=1$. Thus, for $x>2$, our function is:

The Verdict

Now, let us test the junction. For continuity, we check the right-hand limit at $x=2$:

Since the left-hand limit was also $11$, the function is continuous! The bridge is intact.

But is it smooth? We check the right-hand derivative:

At $x=2$, $f^{\prime }(2^{+})=2+4=6$. Our left-hand derivative was $5$.

Since $5 eq 6$, the slopes do not match. We have discovered a sharp corner! The function is continuous, but not differentiable at $x=2$.