Answer:
Option D is correct.
The domain of the function f(x) is all real numbers except 2 and 4.
Step-by-step explanation:
f(x) = (x+1)/(x²-6x+8)
The domain of a function expresses the region of values of x, where the function exists.
And logically, a function exists where ever f(x) has a finite value. That is, the only point where A function does not exist is when f(x) gives infinity.
For a rational function, the point where a function doesn't exist is when the denominator of the rational function is equal to 0. Because (numerator/0) --> ∞
So, the denominator in this question is
x²-6x+8
The function doesn't exist when
x²-6x+8 = 0
So, we solve the quadratic equation that ensues to get the values of x where the function doesn't exist.
x²-6x+8 = 0
x² - 4x - 2x + 8 = 0
x(x-4) - 2(x-4) = 0
(x-2)(x-4) = 0
(x-2) = 0 or (x-4) = 0
x = 2 or x = 4
This means that the function doesnt exist at x = 2 and x = 4
Indicating further that the function exists everywhere except at x = 2 and x = 4.
Hence, from the definition of domain given above, it is clear that the domain of the given function is all real numbers except 2 and 4.
Hope this Helps!!!
