Answer:
The domain of the function is all real values of x, except [tex]x = 4[/tex] and [tex]x = 2[/tex]
Step-by-step explanation:
We are given the following function:
[tex]f(x) = \frac{x+1}{x^2-6x+8}[/tex]
It's a fraction, so the domain is all the real values except those in which the denominator is 0.
Denominator:
Quadratic equation with [tex]a = 1, b = -6, c = 8[/tex]
Using bhaskara, the denominator is 0 for these following values of x:
[tex]\Delta = (-6)^2 - 4(1)(8) = 36-32 = 4[/tex]
[tex]x_{1} = \frac{-(-6) + \sqrt{4}}{2} = 4[/tex]
[tex]x_{2} = \frac{-(-6) - \sqrt{4}}{2} = 2[/tex]
The domain of the function is all real values of x, except [tex]x = 4[/tex] and [tex]x = 2[/tex]
