Animated Solution for Mathematics - Functions: The domain of the function $f(x)=\frac{{\sin }^{-1}(x-3)}{\sqrt{9-x^2}}$ is

To determine the domain of the function $f(x)=\frac{{\sin }^{-1}(x-3)}{\sqrt{9-x^2}}$, we must ensure that every component of the expression is mathematically valid. The function remains defined only when all individual constraints are satisfied simultaneously.

If any part of the function fails—such as taking the square root of a negative number or dividing by zero—the function ceases to exist. We must find the intersection of all valid intervals.

The inverse sine function, ${\sin }^{-1}(u)$, is defined only for inputs $u$ such that $-1\le u\le 1$. For our numerator, we require:

To isolate $x$, we add $3$ to all parts of the inequality. This yields:

Thus, the numerator is defined for $x\in [2,4]$. This represents our first safety zone.

The denominator $\sqrt{9-x^2}$ imposes two strict conditions. First, the radicand must be non-negative ($9-x^2\ge 0$), and second, the denominator cannot be zero ($9 - x^2 eq 0$). Combining these, we require:

Rearranging this inequality, we obtain:

This simplifies to the absolute value inequality $|x|<3$, which corresponds to the open interval $(-3,3)$. This is our second safety zone.

The function is valid only where both safety zones overlap. We must find the intersection of the intervals $[2,4]$ and $(-3,3)$.

Visualizing this on a number line, the interval starts at $x=2$ (which is included in both sets). The overlap continues until $x=3$.

Since $x=3$ is excluded from the second interval (the denominator would become zero), it must also be excluded from our final result.

By intersecting these regions, we find that the function is defined for all $x$ such that $2\le x<3$.

The final domain of the function is: