For what real values of $a$ does the range of $f(x)$ contains the interval $[0,1]$?

Question

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.

Answer 1

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 $$ a \le 0 \implies range\left(\frac{x+1}{x^2 + a}\right) \subseteq[0,1]. $$

Answer 2

Here what I have come up, but you better analyze deeply !!! $$0\leq \frac{x+1}{a+x^2}\leq 1$$

You may have three possibilities which depend on $x$:

If $x<-1$ then $a\leq -x^2+x+1$.

If $x=-1$ then $(a<-1\lor a>-1)$, i.e., $a\neq -1$.

If $x>-1$ then $a\geq -x^2+x+1$.

I think that you cannot exactly specify the value of $a$ without $x$.

Answer 3

Up front I will say that I did not leverage the suggestions from your book. But here is what my analysis yields regarding values for $a$.

When $a=-1$ the rational function $y$ reduces to $\frac{1}{x-1}$ which never takes on the value of zero. So one immediate restriction is $a \ne -1$.

Assuming the above restriction, set $y=1$ and solve for $x$ which gives: $$x = \frac {1 \pm \sqrt{5-4a}}{2}.$$

In order for $y$ to ever achieve a value of $1$ we then need $a \le 5/4$. In fact values of $a$ that exceed this bound will yield values of $y$ that will always be less than $1$.

When $0 < a \le 5/4$ the function smoothly passes through the range $[0, 1]$ without any drama.

At $a=0$ we pick up a single positive verticle asymptote at $x=0$ yet the function will still pass through the desired range as $x$ approaches $0$ from below.

When $-1<a<0$ there will be a pair of vertical asymptotes and yet the function will only pass through the range $[0, 1]$ as $x$ approaches $-\sqrt {|a|}$ from below.

When $a < -1$ the function will pass through the range $(-\infty, \infty)$ and hence the range of interest when $x$ falls in $(-\sqrt {|a|}, \sqrt {|a|})$.

So it appears that the function will pass through the range $[0, 1]$ when $a \le 5/4$ and $a \ne -1$.