For this case we have that by definition, the domain of a function, is given for all the values for which the function is defined.
We have:
![f (x) = \frac {x + 1} {x ^ 2-6x + 8}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/7ttnbgbwz1jwk1wh8918ljuo4d3q1c8d5g.png)
The given function is not defined when the denominator is equal to zero. That is to say:
![x ^ 2-6x + 8 = 0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/q9jfp5dz112mqg0xd7q08l4sn05thhsan2.png)
To find the roots we factor, we look for two numbers that when multiplied give as a result "8" and when added as a result "-6". These numbers are:
![-4-2 = -6\\-4 * -2 = 8](https://img.qammunity.org/2020/formulas/mathematics/middle-school/kdk2m14vap8r7oh4vhxpwg6vkzjnkv3hzu.png)
Thus, the factored polynomial is:
![(x-4) (x-2) = 0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/6sqp0ldiwfbhn5iiy9c6ia3jdse4k4neja.png)
That is to say:
![x_ {1} = 4\\x_ {2} = 2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/qdgwarqrsj5kkmvb008qcna6yymfxifcw4.png)
Makes the denominator of the function 0.
Then the domain is given by:
All real numbers, except 2 and 4.
Answer:
x |x≠2,4