Question

find the domain of the function f. use limits to describe the behavior of f at value(s) of x not in its domain
f(x) = \frac{1}{x + 6}

select the correct choice for the domain below and, if necessary, fill in the answer box to complete your answer
a. the domain is the set of all real numbers except x ≠ -6
(use a comma to separate answers as needed.)
b. the domain is the set of all real numbers.
\lim_{x \to -6^-} f(x) = \square (simplify your answer.)
\lim_{x \to -6^+} f(x) = \square (simplify your answer.)

Explanation:

Step1: Analyze the function's domain

The function \( f(x)=\frac{1}{x + 6} \) is a rational function. A rational function is undefined when the denominator is zero. So, we set the denominator equal to zero and solve for \( x \):
\( x+6 = 0 \)
\( x=-6 \)
Thus, the domain of \( f(x) \) is all real numbers except \( x = - 6 \), which matches option A.

Step2: Find the left - hand limit as \( x

ightarrow - 6^{-} \)
As \( x
ightarrow - 6^{-} \) (x approaches - 6 from the left), we consider values of \( x \) that are less than - 6, say \( x=-6 - h \) where \( h>0 \) and \( h
ightarrow0 \).
Substitute \( x=-6 - h \) into the function \( f(x)=\frac{1}{x + 6} \), we get \( f(-6 - h)=\frac{1}{(-6 - h)+6}=\frac{1}{-h} \).
As \( h
ightarrow0^{+} \) (since \( x
ightarrow - 6^{-} \) implies \( h
ightarrow0^{+} \)), \( \frac{1}{-h}
ightarrow-\infty \) because the numerator is 1 (a positive constant) and the denominator is a negative number that approaches 0. So, \( \lim_{x
ightarrow - 6^{-}}f(x)=-\infty \).

Step3: Find the right - hand limit as \( x

ightarrow - 6^{+} \)
As \( x
ightarrow - 6^{+} \) (x approaches - 6 from the right), we consider values of \( x \) that are greater than - 6, say \( x=-6 + h \) where \( h>0 \) and \( h
ightarrow0 \).
Substitute \( x=-6 + h \) into the function \( f(x)=\frac{1}{x + 6} \), we get \( f(-6 + h)=\frac{1}{(-6 + h)+6}=\frac{1}{h} \).
As \( h
ightarrow0^{+} \) (since \( x
ightarrow - 6^{+} \) implies \( h
ightarrow0^{+} \)), \( \frac{1}{h}
ightarrow+\infty \) because the numerator is 1 (a positive constant) and the denominator is a positive number that approaches 0. So, \( \lim_{x
ightarrow - 6^{+}}f(x)=+\infty \).

Answer:

The domain option: A. The domain is the set of all real numbers except \( x=-6 \)
\( \lim_{x
ightarrow - 6^{-}}f(x)=-\infty \)
\( \lim_{x
ightarrow - 6^{+}}f(x)=+\infty \)