Find $a, b \in \mathbb{R}$ such that $f: \mathbb {R} \rightarrow [-1, 1], \ f(x) = \frac{ax + b}{x^2 + 1}$ is surjective

Question

I found this problem in a book and I cannot solve it. Find $a, b \in \mathbb{R}$ sucht that $f:\mathbb{R} \rightarrow [-1, 1], \ f(x) = \frac{ax + b}{x^2 + 1}$ is surjective.

I have tried to do it the usual way: for every $y \in [-1, 1]$, there is a $x \in \mathbb{R}$ such that $f(x) = \frac{ax + b}{x^2 + 1} = y$ and solved the equation: $$x = \frac{a \pm \sqrt {a^2 - 4y^2 + 4b}}{2y}$$ Now, for every $x \in \mathbb{R}$, it exists a $y\in \mathbb{R}$, such that $\frac{a \pm \sqrt{a^2 - 4y^2 + 4b}}{2y}$, but I could not continue with this idea.

Another idea was set $f(x) \ge -1$ and $f(x) \le 1$. I get 2 quadratics to be greater than $0$, which means $\Delta \le 0$ and such I get a system of 2 quadratic inequations in $a$. But this does not guarantee the fact that the whole interval $[-1, 1]$ is covered.

Have you got any idea?

Answer 1

If $a=0$, no $b$ exists such that $f(\Bbb R)=[-1,1]$, because if $b\geqslant0$, then $(\forall x\in\Bbb R):f(x)\geqslant0$ and if $b\leqslant0$, then $(\forall x\in\Bbb R):f(x)\leqslant0$.

Now, suppose that $a>0$. Note that$$f'(x)=\frac{-a x^2-2 b x+a}{\left(x^2+1\right)^2}$$and that therefore$$f'(x)=0\iff x=\frac{-b\pm\sqrt{a^2+b^2}}a.$$So, since $a>0$, $f$ is decreasing on $\left(-\infty,\frac{-b-\sqrt{a^2+b^2}}a\right]$ and on $\left[\frac{-b+\sqrt{a^2+b^2}}a,\infty\right)$ and increasing on $\left[\frac{-b-\sqrt{a^2+b^2}}a,\frac{-b+\sqrt{a^2+b^2}}a\right]$. But$$f\left(\frac{-b-\sqrt{a^2+b^2}}a\right)=\frac{b-\sqrt{a^2+b^2}}2\quad\text{and}\quad f\left(\frac{-b+\sqrt{a^2+b^2}}a\right)=\frac{b+\sqrt{a^2+b^2}}2.$$So, you must have$$\frac{b-\sqrt{a^2+b^2}}2=-1\quad\text{and}\quad\frac{b+\sqrt{a^2+b^2}}2=1.$$If you add these equalities, you get that $b=0$. It follows now that $a=2$.

The case in which $a<0$ is similar.

Answer 2

We know the $f'$ has at most two zeroes that is at most two extrema.. Now the maximal value of $f$ is $1$ and its minimal value is $-1$, hence there are exactly two extreme values.

Solving $f(x)=1$ gives
$$x=\frac12\left(a\pm\sqrt{a^2+4b-4}\right).$$ But as the maximum value is $1$ it must be attained only once, hence $\sqrt{a^2+4b-4}$ must be zero, that is $$a^2+4b-4=4.$$ Similarly, solving for $f(x)=-1$, we obtain $$a^2-4b-4=4.$$ Immediately $a=\pm2$ and $b=0$ follow.

Answer 3

Oh... here's a thing. You nee there to be $x$ so that $\frac {ax+b}{x^2 + 1} = 1$ but there can't be any $x$ so that $\frac {ax+b}{x^2 + 1} > 1$.

To solve $\frac {ax+b}{x^2 + 1} = 1$ then we have $x^2 - ax +(1-b) = 0$ and $x =\frac {a\pm \sqrt{a^2 - 4(1-b)}}2$

So $a^2 \ge 4(1-b)$.

But for any $1+ \epsilon > 1$ we have $\frac {ax+b}{x^2 + 1} = 1+\epsilon$ or

$ax+b = x^2(1+\epsilon) + (1+ \epsilon)$

$x = \frac {a\pm \sqrt{a^2 - 4(1+\epsilon-b)}}{2(1+\epsilon)}$ cant exist or in other words

$a^2 < 4(1+\epsilon -b)$ for all positive $\epsilon$ but

$a^2 \ge 4(1-b)$.

The only way that is possible is if

$a^2 = 4(1-b)$.

Now is we do the same thing for finding there must be an $x $ so that $\frac {ax+b}{x^2 + 1} = -1$ but there can't be any $x $ so that $\frac {ax+b}{x^2 + 1} < -1$.

Same reasoning we get

$x^2 + ax + (1+b) = 0$ so we must have $a^2 \ge 4(1+b)$ but for any $\epsilon > 0$ we have

$a^2 < 4(1-\epsilon + b)$ and the only way that is possible is if $a^2 = 4(1+b)$.

So $a^2 = 4(1+b) =4(1-b)$

So $b = 0$ and $a= \pm 2$.

...

We must confirm though that all values for $y\in [-1,1]$ are possible.

That is that for all $-1\le y \le 1$ then $a^2 \ge 4y(y-b) =4y^2 - 4yb$

Or $4 \ge 4y^2$. Which is clear and $0\le y^2 \le 1$.

Answer 4

We know (by various familiar means) that the horizontal asymptote of $ \ f(x) \ $ is zero, and that the function is defined and continuous "everywhere" in $ \ \mathbb{R} \ \ , $ so we will be interested in what is necessary to restrict its range to $ \ -1 \ \le \ f(x) \ \le \ +1 \ \ $ for finite values of $ \ x \ \ . $

The function $ \ f(x) \ = \ \frac{ax + b}{x^2 + 1} \ $ can be written as the sum of even and odd functions. The values of the even-symmetry component $ \ \frac{ b}{x^2 + 1} \ $ can only have the same sign as $ \ b \ \ , $ so it cannot cover the desired range alone. The odd-symmetry component $ \ \frac{ ax}{x^2 + 1} \ $ has the derivative $ \ \frac{a·(1 - x^2)}{(x^2 + 1)^2} \ \ , $ so its extrema are located at $ \ x \ = \ \pm 1 \ \ $ for all $ \ a \ \neq 0 \ \ , $ those values being $ \ f(\pm \ 1) \ = \ \pm \frac{a}{2} \ \ . $ So this component can cover the entire interval $ \ [-1 \ , \ 1] \ $ by itself for $ \ \mathbf{a \ = \ \pm 2} \ \ , $ thus leaving $ \ \mathbf{b \ = \ 0} \ \ . $ (The "tails" of the function cannot reach $ \ y = 0 \ \ , $ but that value is "covered" since the curve of an everywhere-continuous, odd-symmetric function must pass through the origin.) So the functions $ \ f(x) \ = \ \pm \frac{2x}{x^2 + 1} \ $ meets our requirement.

In fact, these functions are bijective for $ \ [-1 \ , \ 1 ] \ \rightarrow \ [-1 \ , \ 1] \ \ . $ We find that $$ \frac{2x}{x^2 + 1} \ \ = \ \ \frac{2x'}{x'^2 + 1} \ \ \Rightarrow \ \ x·x'^2 + x \ \ = \ \ x'·x^2 + x' \ \ \Rightarrow \ \ x·x'·(x' - x) \ \ = \ \ x' \ - \ x \ \ ; $$ so for $ \ x·x' \ \neq \ 0 \ \ , \ x' \ = \ x \ \ . $ The inverse function to $ \ \frac{2x}{x^2 + 1} \ $ is obtained from $ \ y·x^2 - 2x + y \ = \ 0 \ \rightarrow \ x \ = \ \frac{1}{y} \ - \ \frac{\sqrt{1 \ - \ y^2}}{y} \ \ $ after choosing signs appropriately; note that $ \ \lim_{y \ \rightarrow \ 0} \ \frac{1 \ - \ \sqrt{1 \ - \ y^2}}{y} \ = \ 0 \ \ . $ Since $ \ f'(x) \ = \ \frac{2·(1 - x^2)}{(x^2 + 1)^2} \ > \ 0 \ \ $ on $ \ (-1 \ , \ 1) \ \ , $ this inverse function is also always increasing on the interval.

The general asymmetric function $ \ f(x) \ = \ \frac{ax + b}{x^2 + 1} \ \ $ has been discussed by the other posters. The derivative function becomes $ \ \frac{-a·x^2 \ - \ 2bx \ + \ a }{(x^2 + 1)^2} \ \ , $ placing the function extrema at $ \ x \ = \ -\frac{b}{a} \ \pm \ \frac{\sqrt{a^2 \ + \ b^2}}{a} \ \ . $ We may then calculate the global maximum and minimum of the function as $$ f \left(-\frac{b}{a} \pm \frac{\sqrt{a^2 \ + \ b^2}}{a} \right) \ \ = \ \ \frac{a·\left(-\frac{b}{a} \pm \frac{\sqrt{a^2 \ + \ b^2}}{a} \right) + b}{\left(-\frac{b}{a} \pm \frac{\sqrt{a^2 \ + \ b^2}}{a} \right)^2 + 1} \ · \ \frac{a^2}{a^2} $$ $$ = \ \ \pm \frac{a^2· \sqrt{a^2 + b^2}} {b^2 \ \pm \ 2·b·\sqrt{a^2 + b^2} \ + \ (a^2 + b^2) \ + \ a^2} \ \ = \ \ \pm \frac{a^2}{2} · \frac{\sqrt{a^2 + b^2}} {(a^2 + b^2) \ \pm \ b·\sqrt{a^2 + b^2 }} $$ $$ = \ \ \pm \frac{a^2}{2} · \frac{1} {\sqrt{a^2 + b^2} \ \pm \ b} \ · \ \frac{\sqrt{a^2 + b^2} \ \mp \ b}{\sqrt{a^2 + b^2} \ \mp \ b} \ \ = \ \ \pm \frac{a^2}{2} · \frac{\sqrt{a^2 + b^2} \ \mp \ b} {a^2 \ + \ b^2 \ - \ b^2} $$ $$ = \ \ \frac{b}{2} \ \pm \ \frac{\sqrt{a^2 + b^2}}{2} \ \ . $$

For $ \ a \ , \ b \ > \ 0 \ \ , $ we then have $$ \frac{b}{2} \ - \ \frac{\sqrt{a^2 + b^2}}{2} \ \ < \ \ 0 \ \ < \ \ b \ \ < \ \ \frac{b}{2} \ + \ \frac{\sqrt{a^2 + b^2}}{2} \ \ . $$ It is not possible here to choose values for $ \ a \ $ and $ \ b \ $ that will simultaneously produce a maximum of $ \ +1 \ $ and a minimum of $ \ -1 \ \ . $ If we "force" the maximum value to be $ \ \frac{b}{2} \ + \ \frac{\sqrt{a^2 \ + \ b^2}}{2} \ = \ +1 $ $ \Rightarrow \ \sqrt{a^2 \ + \ b^2} \ = \ 2 - b \ > \ 0 \ , $ then the minimum value is $ \ \frac{b \ - \ (2 - b)}{2} \ = \ 1 - b \ > \ -1 \ \ . $ Alternatively, setting the minimum value of $ \ \frac{b}{2} \ - \ \frac{\sqrt{a^2 \ + \ b^2}}{2} \ = \ -1 \ \Rightarrow \ \sqrt{a^2 \ + \ b^2} \ = \ b + 2 \ ; \ $ since $ \ b > 0 \ \ , $ we have $ \ \sqrt{a^2 \ + \ b^2} \ = \ b + 2 \ > \ 2 \ \ , \ $ which drives the maximum of the function to be $ \ \frac{b \ + \ (b + 2)}{2} \ = \ b + 1 \ > \ 1 \ \ . $ Either the range fails to "cover" $ \ [-1 \ , \ 1] \ $ completely, or it extends beyond that interval. (Similar situations arise for the other choices of signs for $ \ a \neq 0 \ , \ b \neq 0 \ \ . \ ) $

So the only functions satisfying the specified conditions are $ \ f(x) \ = \ \pm \frac{2x}{x^2 + 1} \ \ . $