Animated Solution for Mathematics - Functions: The domain of the function $f(x)=\frac{1}{\sqrt{[x{]}^2-3[x]-10}}$ is (where $[x]$ denotes the greatest integer less than or equal to $x$)

Welcome, fellow traveler, to the fascinating world of function domains! Today, we are going to dissect a problem that seems simple on the surface but hides a beautiful, logical trap.

We are looking at the function:

At first glance, it looks like a standard algebra problem, but it is actually a test of your understanding of the Greatest Integer Function (GIF). Let us peel back the layers together.

The first thing we must do is ensure our function is mathematically sound. We have a square root in the denominator.

Normally, the expression inside a square root must be greater than or equal to zero. However, because this square root is sitting in the denominator, we have an additional constraint: the denominator cannot be zero.

Therefore, the expression inside the square root must be strictly greater than zero:

This is our golden rule for this problem.

Dealing with the GIF directly can be intimidating. Let us simplify our lives by introducing a substitution. Let $t=[x]$.

Now, our inequality transforms into a familiar quadratic:

We are looking for two numbers that multiply to $-10$ and add to $-3$. Those numbers are $-5$ and $2$. Thus, we can factorize the inequality as:

Using the wavy curve method, we identify the critical points at $t=-2$ and $t=5$. For the product to be positive, $t$ must lie in the outer regions: $t<-2$ or $t>5$.

Now, we must return to our original variable, $x$. We have two cases:

Case 1: $[x]<-2$. Since $[x]$ must be an integer, the largest integer strictly less than $-2$ is $-3$. So, we require $[x]\le -3$. By the definition of the GIF, if $[x]\le -3$, then $x$ must be strictly less than $-2$. This gives us the interval $(-\infty ,-2)$.

Case 2: $[x]>5$. The smallest integer strictly greater than $5$ is $6$. So, we require $[x]\ge 6$. By the definition of the GIF, if $[x]\ge 6$, then $x$ must be greater than or equal to $6$. This gives us the interval $[6,\infty )$.

Finally, we combine these two valid regions using the union operator.

Our domain is:

It is a beautiful result, isn't it? The key takeaway here is that when you work with the greatest integer function, you must always think about the integer boundaries.

Never rush the transition from the integer $t$ back to the real number $x$. Keep practicing, keep visualizing, and you will master these concepts in no time!