Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $[x]$ be the greatest integer $\le x$. Then the number of points in the interval $(-2,1)$ where the function $f(x)=|[x]|+\sqrt{x-[x]}$ is discontinuous, is _____.

Welcome, future engineers! Today, we are going to dissect a function that looks intimidating at first glance but reveals a beautiful, rhythmic structure once we peel back the layers. We are dealing with $f(x)=|[x]|+\sqrt{x-[x]}$.

At first, the Greatest Integer Function $[x]$ might make you nervous. It is notorious for creating 'jumps' in graphs. But remember, in mathematics, fear is just a lack of information. Let us transform that fear into understanding.

First, let us look at the term under the square root: $x-[x]$. Does this look familiar? It is the definition of the fractional part of $x$, denoted as $\{x\}$.

So, our function is actually $f(x)=|[x]|+\sqrt{\{x\}}$. This is much cleaner!

Now, we are looking for discontinuities in the open interval $(-2,1)$. Because the Greatest Integer Function is constant between integers, the function will behave like a simple algebraic curve in those intervals, but it will 'jump' whenever $x$ crosses an integer. This is where our investigation begins.

To analyze this, we must break our interval $(-2,1)$ into three distinct zones where $[x]$ remains constant:

1. For $x\in (-2,-1)$, the value of $[x]$ is $-2$. Thus,

2. For $x\in [-1,0)$, the value of $[x]$ is $-1$. Thus,

3. For $x\in [0,1)$, the value of $[x]$ is $0$. Thus,

Imagine these as three separate tracks on a racecourse. Within each track, the function is smooth. The danger—and the excitement—happens at the transition points: $x=-1$ and $x=0$.

Now, we test the continuity at these transitions. Continuity requires the Left Hand Limit (LHL) to equal the Right Hand Limit (RHL).

At $x=-1$: - LHL: ${\lim }_{x\to -1^{-}}(2+\sqrt{x+2})=2+\sqrt{1}=3$. - RHL: ${\lim }_{x\to -1^{+}}(1+\sqrt{x+1})=1+\sqrt{0}=1$.

Since $3 eq 1$, the function has a jump discontinuity at $x=-1$. The graph literally breaks here.

At $x=0$: - LHL: ${\lim }_{x\to 0^{-}}(1+\sqrt{x+1})=1+\sqrt{1}=2$. - RHL: ${\lim }_{x\to 0^{+}}\sqrt{x}=0$.

Since $2 eq 0$, we have another jump discontinuity at $x=0$.

By systematically checking the boundaries, we have identified exactly two points of discontinuity: $x=-1$ and $x=0$.

The total number of points of discontinuity is 2.

This problem teaches us a vital lesson: never be intimidated by complex-looking functions. Break them down into their constituent parts, analyze the behavior in the 'safe' zones, and rigorously test the 'danger' zones. You have mastered the logic; now go forth and apply this to every problem you face!