Question

what is the domain of the function $f(x)=\frac{x + 1}{x^2 - 8x + 8}$?

all real numbers

all real numbers except $-1$

all real numbers except $-4$ and $-2$

all real numbers except $2$ and $4$

Explanation:

Step1: Identify the denominator

The function is \( f(x) = \frac{x + 1}{x^2 - 8x + 8} \)? Wait, there's a typo? Wait, maybe it's \( x^2 - 6x + 8 \)? Wait, the options have 2 and 4. Let's check: \( x^2 - 6x + 8 \) factors to \( (x - 2)(x - 4) \). Wait, maybe the original denominator is \( x^2 - 6x + 8 \). Let's assume that (since the options have 2 and 4). So, the denominator is \( x^2 - 6x + 8 \).

Step2: Set denominator equal to zero

To find the values that make the denominator zero (which are excluded from the domain), we solve \( x^2 - 6x + 8 = 0 \).

Step3: Factor the quadratic

Factor \( x^2 - 6x + 8 \): we need two numbers that multiply to 8 and add to -6. Those numbers are -2 and -4. So, \( x^2 - 6x + 8 = (x - 2)(x - 4) \).

Step4: Solve for x

Set each factor equal to zero:

• \( x - 2 = 0 \) gives \( x = 2 \)
• \( x - 4 = 0 \) gives \( x = 4 \)

So, the denominator is zero when \( x = 2 \) or \( x = 4 \). Therefore, the domain is all real numbers except 2 and 4.

Answer:

D. all real numbers except 2 and 4 (assuming the options are labeled A to D as: A. all real numbers, B. all real numbers except -1, C. all real numbers except -4 and -2, D. all real numbers except 2 and 4)