If $y = \frac{x + 1}{x^2 + 1 - x},$ and $x$ is any real number, then what is the sum of the maximum and minimum possible values of $y \, ?$
First, we simplify the expression:
$$y = \frac{x + 1}{x^2 + 1 - x} = \frac{x + 1}{x^2 - x + 1}.$$
In turn, we decompose the denominator:
$$x^2 - x + 1 = (x - 1)^2 + x^2 - 2x + 1 = (x - 1)^2 + (x - 1) + 2 = (x - 1)\left[ x - 1 + 2 \cdot \frac{1}{x - 1}\right].$$To find the minimum and maximum possible values of $y,$ we consider the numerator and denominator separately:
The numerator is $x + 1$. Since $x$ is real,
the maximum value of $x + 1$ is 2, which occurs when $x = 1,$ and
the minimum value of $x + 1$ is $-\infty,$ which occurs when $x \to -\infty.$
The denominator is $(x - 1)(x - 1 + 2/(x - 1)).$ In turn,
the expression $(x - 1 + 2/(x - 1))$ is increasing on $[1,\infty)$ and hence has a minimum value of $x = 1,$ which is 2.
Thus, $(x - 1)(x - 1 + 2/(x - 1))$ is negative when $x < 1,$ positive when $x > 1.$
Therefore, $y$ is negative when $x < 1,$ and positive when $x > 1.$
To put everything together,
the minimum possible value of $y$ is $\boxed{-\infty},$
the maximum possible value of $y$ is 2,
and our answer is $-\infty + 2 = \boxed{-\infty}.$
To find the sum of the maximum and minimum possible values of $y,$ we need to find the range (or possible values) of $y.$
Given that $x$ is any real number, we can start by examining the denominator $x^2 + 1 - x.$
Factorize the denominator as $x^2 - x + 1.$
To find the range of $x$ for which the denominator is positive, we analyze the discriminant $b^2 - 4ac$ of the quadratic equation $x^2 - x + 1 = 0.$
The discriminant is given by $(-1)^2 - 4(1)(1) = -3.$
Since the discriminant is negative, the quadratic equation $x^2 - x + 1 = 0$ has no real solutions.
This means the quadratic $x^2 - x + 1$ is always positive, no matter the value of $x.$
Therefore, the denominator $(x^2 + 1 - x)$ is positive for all real values of $x.$
Now let's analyze the numerator $x + 1.$
Since $x$ can take any real value, the numerator can be any real number $+1.$
Hence, the possible values of $y = \frac{x + 1}{x^2 + 1 - x}$ are all real numbers.
Therefore, the sum of the maximum and minimum possible values of $y$ is $+\infty + (-\infty) = \boxed{\text{undefined}}.$