To determine the range of the function [tex]\( f(x) = \frac{1}{x + 3} \)[/tex], let's analyze its behavior step-by-step.
1. Identify where the function is undefined:
- The function [tex]\( f(x) = \frac{1}{x + 3} \)[/tex] involves division. Division by zero is undefined in mathematics.
- Set the denominator equal to zero to find the points where the function is undefined:
[tex]\[
x + 3 = 0 \implies x = -3
\][/tex]
- Therefore, the function [tex]\( f(x) = \frac{1}{x + 3} \)[/tex] is undefined when [tex]\( x = -3 \)[/tex].
2. Determine the behavior of the function at values approaching the undefined point:
- As [tex]\( x \)[/tex] approaches [tex]\(-3\)[/tex] from the left (i.e., [tex]\( x \to -3^- \)[/tex]), [tex]\( x + 3 \)[/tex] approaches zero from negative values. As a result, [tex]\( \frac{1}{x + 3} \)[/tex] becomes a very large negative number.
- As [tex]\( x \)[/tex] approaches [tex]\(-3\)[/tex] from the right (i.e., [tex]\( x \to -3^+ \)[/tex]), [tex]\( x + 3 \)[/tex] approaches zero from positive values. Consequently, [tex]\( \frac{1}{x + 3} \)[/tex] becomes a very large positive number.
3. Analyze the limits and behavior for [tex]\( x \)[/tex] not equal to [tex]\(-3\)[/tex]:
- For any other value of [tex]\( x \)[/tex] besides [tex]\(-3\)[/tex], [tex]\( x + 3 \)[/tex] is not zero, and the function [tex]\( f(x) \)[/tex] is defined.
- As [tex]\( x \)[/tex] moves toward positive or negative infinity, [tex]\( f(x) \)[/tex] approaches zero but never actually reaches zero.
4. Determine the complete range:
- This means [tex]\( f(x) = \frac{1}{x + 3} \)[/tex] can take any real value except zero.
- Therefore, the range of [tex]\( f(x) \)[/tex] is all real numbers except zero.
In interval notation, this range is expressed as:
[tex]\[
(-\infty, 0) \cup (0, \infty)
\][/tex]
Hence, the correct answer is:
A. [tex]\( (-\infty, 0) \cup (0, \infty) \)[/tex]
