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 $-\infty + 2 = \boxed{-\infty}.$
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 $y \, ?$
1 answer