Question

determine the point(s) at which the given function f(x) is continuous. f(x) = 8 csc (9x) the function is continuous on (-∞,∞) except for \\(\frac{npi}{9}\\) (type an exact answer, using \\(pi\\) as needed. type an expression using n, where n is an integ

Explanation:

Step1: Recall the definition of cosecant function

The cosecant function is defined as \(\csc(\theta)=\frac{1}{\sin(\theta)}\), so \(f(x) = 8\csc(9x)=\frac{8}{\sin(9x)}\). A function is undefined (and thus not continuous) where its denominator is zero. So we need to find where \(\sin(9x) = 0\).

Step2: Solve \(\sin(9x)=0\)

We know that \(\sin(\theta)=0\) when \(\theta = n\pi\), where \(n\) is an integer. Let \(\theta = 9x\), so \(9x=n\pi\).

Step3: Solve for \(x\)

Divide both sides of \(9x = n\pi\) by 9: \(x=\frac{n\pi}{9}\), where \(n\in\mathbb{Z}\) (the set of all integers). So the function \(f(x)\) is continuous everywhere except at \(x = \frac{n\pi}{9}\) for all integers \(n\).

Answer:

The function \(f(x)=8\csc(9x)\) is continuous on \((-\infty,\infty)\) except at \(x = \frac{n\pi}{9}\) where \(n\) is an integer.