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Discontinuity in a function occurs when there is a break, jump, or hole in the graph of the function. For a function to be continuous at a point, three conditions must be satisfied:

• The function must be defined at the point.
• The limit of the function as it approaches the point must exist.
• The limit of the function must equal the function's value at that point.

In the function \(f(x)=\frac{\sqrt{\tan^{-1}x}}{x^2-9}\), potential discontinuities occur when the denominator is zero, which happens at \(x=3\) and \(x=-3\). This is because dividing by zero is undefined in mathematics.

So, \(f(x)\) is discontinuous at \(x=3\) and \(x=-3\). At these points, we might find vertical asymptotes, which indicate a break in the continuity of the function. Recall that a key step in analyzing continuity is identifying where the denominator of rational functions equals zero.

The function in question here, \(f(x) = \frac{\sqrt{\tan^{-1} x}}{x^2 - 9}\), is an example of a composite function. A composite function is made by using one function as an input for another. In this case, the numerator \(\sqrt{\tan^{-1} x}\) is a composite function.

Breaking it down:

• The innermost function is \(\tan^{-1} x\), which is the inverse tangent function. It takes any real number and returns an angle between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).
• The outer function is the square root function \(\sqrt{x}\), which is defined when its input is non-negative.

Since \(\tan^{-1} x\) is always non-negative for all real \(x\), the function \(\sqrt{\tan^{-1} x}\) is continuous over the entire real line. By understanding each component and their domains, we can effectively analyze the continuity of the composite function.

To find intervals of continuity for \(f(x) = \frac{\sqrt{\tan^{-1} x}}{x^2 - 9}\), we need to consider where the function is defined and continuous. First, ensure the numerator, \(\sqrt{\tan^{-1} x}\), is valid, which it is for all real \(x\).

The challenge lies in the denominator, \(x^2 - 9\), which cannot be zero. Therefore, \(x\) should not be 3 or -3. These values divide the real number line into three distinct segments:

• \(( -\infty, -3)\)
• \((-3, 3)\)
• \((3, \infty)\)

In each interval, the function is continuous because both the numerator is well-defined and the denominator does not pose any issue. Thus, \(f(x)\) is continuous everywhere except at \(x=3\) and \(x=-3\), creating these specific intervals of continuity. Always verify the overall function, by analyzing both the numerator's and denominator's roles in defining its continuity.