Question

determine where the function f(x) = \frac{3x}{x^2 + 9} is continuous. the function is continuous on \square. (simplify your answer. type your answer in interval notation.)

Explanation:

Step1: Analyze the denominator

To find where the function \( F(x)=\frac{3x}{x^{2}+9} \) is continuous, we first check the denominator \( x^{2}+9 \). We need to see if there are any real values of \( x \) that make the denominator zero, because a function is undefined (and thus discontinuous) where its denominator is zero.
Set the denominator equal to zero: \( x^{2}+9 = 0 \).

Step2: Solve for \( x \)

For the equation \( x^{2}+9 = 0 \), we can rewrite it as \( x^{2}=- 9 \). But for real numbers, the square of a real number \( x \) (i.e., \( x^{2} \)) is always non - negative ( \( x^{2}\geq0 \) for all real \( x \)). So, \( x^{2}=-9 \) has no real solutions. This means that the denominator \( x^{2}+9 \) is never zero for any real number \( x \).

Step3: Determine the domain of continuity

Since the function \( F(x)=\frac{3x}{x^{2}+9} \) is a rational function (a ratio of two polynomials), and the denominator is never zero for real \( x \), the function is defined for all real numbers. In interval notation, the set of all real numbers is \( (-\infty,\infty) \).

Answer:

\((-\infty, \infty)\)