Animated Solution for Mathematics - Inverse Trigonometric Functions: The domain of the function $f(x)={\sin }^{-1}({\log }_2\frac{x^2}{2})$ is given by .........

Welcome, fellow traveler of the mathematical landscape. Today, we are going to tackle a problem that might look like a simple algebraic exercise, but it is actually a beautiful dance of constraints.

We are looking for the domain of the function $f(x)={\sin }^{-1}({\log }_2(\frac{x^2}{2}))$. In the world of functions, the domain is the VIP list—the set of all inputs that are allowed to enter the function without causing a mathematical catastrophe.

Imagine the function as a series of nested rooms. To get to the center, you must pass through the gatekeepers of each layer. Our first gatekeeper is the ${\sin }^{-1}$ function.

The ${\sin }^{-1}$ Constraint

The outermost layer is ${\sin }^{-1}(u)$. The fundamental rule for the inverse sine function is that it only accepts inputs $u$ that live in the closed interval $[-1,1]$.

If you try to feed it anything less than $-1$ or greater than $1$, the function simply refuses to exist. So, our first step is to set the argument of the sine inverse—which is ${\log }_2(\frac{x^2}{2})$—to be trapped within this safe zone.

We write this as:

This is our primary inequality, the bedrock upon which our solution is built.

Now, we need to peel back the next layer: the logarithm. We have ${\log }_2(\frac{x^2}{2})$ trapped between $-1$ and $1$.

To isolate the argument $\frac{x^2}{2}$, we perform the inverse operation of the logarithm, which is exponentiation. We raise all parts of the inequality as powers of the base, which is $2$.

Since our base is $2$, which is greater than $1$, the function is strictly increasing, and the inequality signs remain exactly as they are:

Simplifying this, we get:

Notice how the logarithm has vanished, leaving us with a much friendlier algebraic expression.

To clear the denominator, we multiply the entire inequality by $2$. This gives us:

This is a compound inequality, meaning both parts must be true simultaneously. We can split this into two conditions: Condition A, where $x^2\ge 1$, and Condition B, where $x^2\le 4$.

For Condition A, $x^2\ge 1$ implies that the absolute value of $x$ must be at least $1$. This means $x$ can be any value greater than or equal to $1$, or any value less than or equal to $-1$. In interval notation, this is $(-\infty ,-1]\cup [1,\infty )$.

For Condition B, $x^2\le 4$ implies that the absolute value of $x$ must be at most $2$. This traps $x$ between $-2$ and $2$, or $[-2,2]$.

We have our two conditions. To find the domain, we need the intersection of these two sets—the values of $x$ that satisfy both conditions at the same time.

If we visualize this on a number line, we are looking for the overlap between the regions $(-\infty ,-1]\cup [1,\infty )$ and $[-2,2]$. The overlap occurs exactly in the intervals $[-2,-1]$ and $[1,2]$.

And there we have it! The domain of our function is:

You have successfully navigated the nested constraints and found the truth hidden within the function. Keep this logical rigor in your toolkit, and no function will ever be able to hide its secrets from you.