Explain, using the theorems, why the function is continuous at every number in its domain. F(x) = 2x2 − x − 6 x2 + 9 F(x) is a polynomial, so it is continuous at every number in its domain. F(x) is a rational function, so it is continuous at every number in its domain. F(x) is a composition of functions that are continuous for all real numbers, so it is continuous at every number in its domain. F(x) is not continuous at every number in its domain. none of these State the domain. (Enter your answer using interval notation.)

F(x) is a rational function with denominator that can never be equal to 0 for all real numbers, so it is continuous at every number in its domain.

Step-by-step explanation:

F(x) = (2x² − x − 6)/(x² + 9)

A continuous function over a given interval/domain, exists everywhere within that imterval/domain.

For functions to be continuous, the function must always exist within the real number domain.

F(x) is an improper polynomial with numerator = (2x² − x − 6) and denominator = (x² + 9). And for polynomials, the range of values x can take on range all over the domain of real numbers, (-∞, ∞).

This expression is also a rational function. For a rational function to be continuous, it must exist everywhere in the domain bing considered (real number domain), that is, the denominator must never be equal to 0 within the domain being considered.

The function given is continuous everywhere in the real number domain because it's denominator is never zero for values of x in the real number domain. F(x) exists everywhere in the real number domain.

The only parts where the function doesn't exist is when the denominator (x² + 9) = 0. And this occurs only in the complex number domain.

x² + 9 = 0

x² = -9

x = ± 3i

So, F(x) is continuous at every number in The real number domain because it is a rational function with a denominator that can never be zero.

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step-by-step explanation:

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