Let's determine the range of the function \( f(x) = \frac{1}{x + 3} \).
1. Identify the domain: First, consider the values of \(x\) for which the function is defined. The function \(f(x) = \frac{1}{x + 3}\) has a denominator of \(x + 3\). The function will be undefined when the denominator is zero, i.e., when \(x + 3 = 0\), or \(x = -3\).
Therefore, the function is defined for all real numbers except \(x = -3\).
2. Express the function in terms of \(y\): To find the range, we need to determine the possible values that \(f(x)\) can take (i.e., the outputs \(y\)). Let's set \(y = f(x)\):
[tex]\[
y = \frac{1}{x + 3}
\][/tex]
Rearrange this equation to solve for \(x\):
[tex]\[
y(x + 3) = 1 \quad \Rightarrow \quad x + 3 = \frac{1}{y} \quad \Rightarrow \quad x = \frac{1}{y} - 3
\][/tex]
3. Analyze the possible values of \(y\): Now, consider the limitations on \(y\):
- If \(y = 0\), the expression \(\frac{1}{y}\) would be undefined because division by zero is not allowed. Therefore, \(y\) cannot be zero.
- For any non-zero value of \(y\), the equation \(x = \frac{1}{y} - 3\) will yield a real number \(x\). Hence, \(y\) can take any real value except zero.
4. Conclusion: Therefore, the range of \(f(x) = \frac{1}{x + 3}\) is all real numbers except zero. In interval notation, this is:
[tex]\[
(-\infty, 0) \cup (0, \infty)
\][/tex]
Thus, the correct answer is:
D. [tex]\((-∞, 0) \cup (0, ∞)\)[/tex]
