Answer:
The domain of the function is all real values of x, except
and
![x = 2](https://img.qammunity.org/2022/formulas/mathematics/college/17rxgdi00d0xuclfg1jq4oqczj0zgbwqpq.png)
Explanation:
We are given the following function:
![f(x) = (x+1)/(x^2-6x+8)](https://img.qammunity.org/2022/formulas/mathematics/college/kk5j9vykzqr1c76girhj0dtzrwg1ru9ihr.png)
It's a fraction, so the domain is all the real values except those in which the denominator is 0.
Denominator:
Quadratic equation with
![a = 1, b = -6, c = 8](https://img.qammunity.org/2022/formulas/mathematics/college/t5dxned8dau2h4vejxp64hm4l9lzi5bhr1.png)
Using bhaskara, the denominator is 0 for these following values of x:
![\Delta = (-6)^2 - 4(1)(8) = 36-32 = 4](https://img.qammunity.org/2022/formulas/mathematics/college/acwpedzu43k5hdxdt3xbw4xwq9j7ptntch.png)
![x_(1) = (-(-6) + √(4))/(2) = 4](https://img.qammunity.org/2022/formulas/mathematics/college/x35ah4kjmg761fvhit5z0fzwbkr1llres2.png)
![x_(2) = (-(-6) - √(4))/(2) = 2](https://img.qammunity.org/2022/formulas/mathematics/college/h6srgypf9uhdajny59sqbjq126jmikxbvh.png)
The domain of the function is all real values of x, except
and
![x = 2](https://img.qammunity.org/2022/formulas/mathematics/college/17rxgdi00d0xuclfg1jq4oqczj0zgbwqpq.png)