Question

select the correct answer.
which function’s domain consists of all real numbers except -2?
a. $f(x) = \frac{1}{x + 2}$
b. $f(x) = 2x$
c. $f(x) = 2x - 2$
d. $f(x) = \frac{1}{\sqrt{x + 2}}$

Explanation:

Step1: Analyze Option A

For the function \( f(x)=\frac{1}{x + 2} \), the denominator cannot be zero. So we set \( x+2
eq0 \), which gives \( x
eq - 2 \). The domain is all real numbers except \( - 2 \).

Step2: Analyze Option B

For the function \( f(x) = 2x \), it is a linear function. There are no restrictions on \( x \) (no denominators, no square roots with restrictions here), so the domain is all real numbers.

Step3: Analyze Option C

For the function \( f(x)=2x - 2 \), it is also a linear function. There are no restrictions on \( x \), so the domain is all real numbers.

Step4: Analyze Option D

For the function \( f(x)=\frac{1}{\sqrt{x + 2}} \), we need two things: the expression inside the square root must be non - negative (\( x + 2\geq0\)) and the denominator (the square root) cannot be zero (\( \sqrt{x + 2}
eq0\)). From \( x + 2\geq0\) and \( \sqrt{x + 2}
eq0\), we get \( x+2>0\), so \( x>-2 \). The domain is all real numbers greater than \( - 2 \).

Answer:

A. \( f(x)=\frac{1}{x + 2} \)