Question :
For what real values of $a$ does the range of $f(x) = \cfrac{x+1}{a+x^2} $ contains the interval $[0,1]$?
My doubt lies in the further preceding of this question.
The book states :
Let $y = \cfrac{x+1}{a+x^2} $ . Which implies – $$yx^2 – x + (ay-1) = 0$$
has real roots for every $\color{blue}{y \in [0,1] }$.
I'm not sure how it concluded that it is real for the given interval of y.
I'm also known to the fact that : $D \ge 0$ for the quadratic to have real roots.
And here, $ D = 1 – 4y(ay-1) \ge 0$
Not sure how to go on from here.
Best Answer
it looks easier to consider the reciprocal $$y = g(x) = \frac{a+x^2}{x+1} = x-1 + \frac {a+1}{x+1}$$ and look for the values $a$ so that the range of $g$ contains $[1, \infty).$
we will break the problem in into three cases: $a > -1, a = -1$ and $ a < -1.$
consider the case $a > -1.$ here we have the graph of $y = g,$ hyperbola with a local min $$ y = 2\sqrt{a+1} - 1 \text{ at } x = -1 + \sqrt{a+1}.$$ for the local min to be less than or equal to $1,$ it is necessary that $$-1 < a \le 0.\tag 1$$
case $a = -1$ we have $g = x-1.$ the range of $g$ certainly contains $[1, \infty)$
in case $a < -1,$ the range of $a$ is $(-\infty, \infty)$ which contains $[1, \infty)$
we can conclude that the constraint on $a$ is $$ a \le 0 \implies range\left(\frac{x+1}{x^2 + a}\right) \subseteq[0,1]. $$