Find $b$ If $f:\mathbb R\to\mathbb R, \;f(x)=\dfrac{x^2+bx+1}{x^2+2x+b}, (b>1)$ and $f(x), \dfrac{1}{f(x)}$ have same bounded Range

Question

Let $f:\mathbb R\to\mathbb R, \;f(x)=\dfrac{x^2+bx+1}{x^2+2x+b}, (b>1)$ and $f(x), \dfrac{1}{f(x)}$ have the same bounded set as their range, then value of $b$ is

My Approach:

Let $f(x)=y=\dfrac{x^2+bx+1}{x^2+2x+b}\implies(y-1)x^2+(2y-b)x+(by-1)=0$

since $x\in \mathbb R$ we must have $D=b^2-4ac\geq 0\implies (4-4b)y^2+4y+(b^2-4)\geq0.......(1)$

Also Let $\dfrac{1}{f(x)}=t=\dfrac{x^2+2x+b}{x^2+bx+1}\implies(t-1)x^2+(tb-2)x+(t-b)=0$

Again since $x\in \mathbb R$ we must have $D=b^2-4ac\geq 0\implies (b^2-4)y^2+4y+(4-4b)\geq0.......(2)$

Since $(1)$ and $(2)$ both are positive for all $ y\in \text{Range}$ we must have $D\leq0\implies(4^2)-4(b^2-4)(4-4b)\leq0 \implies b^3-b^2-4b+5\leq0$.

After plotting graph of $b^3-b^2-4b+5=0$ in Desmos I got $b^3-b^2-4b+5\leq0$ for $b\leq {-2.08}$ which is wrong because it is given that $b>1$.

How can i get value of $b$ more than $1$?

Answer 1

Following on from what you have done, you have the range of $f$, i.e. $y$, satisfying the inequality $$y^2(4-4b)+4y+b^2-4\geq0$$ and you have the range of $\frac1f$, also $y$, satisfying the inequality $$(b^2-4)y^2+4y +(4-4b)\geq0$$

The solutions of these inequalities must be the same set, for a given value of $b$, since the ranges of $f$ and $\frac1f$ are identical. This is only possible of these two quadratics are identical, so therefore $b$ must satisfy $$b^2-4=4-4b$$

Since $b>1$ this leads to $$b=-2+2\sqrt{3}$$

You will then be able to obtain the common range, if desired, by solving the inequalit(ies) for this value of $b$.

Answer 2

This is not an official answer, but explains most of the process.

Note: $$\lim_{x \to \pm\infty}{f(x)} = 1$$ and $$\lim_{x \to \pm\infty}{\dfrac{1}{f(x)}} = 1$$ This can be evaluated through partial fractions.

Observe at $b = 1$, $x + 2x + b = (x + 1)^2$ has zeros and any $b > 1$, it does not. This suggests $f(x)$ will not have asymptotes but global max and min when $f'(x) = 0$, precisely at $x = \dfrac{b - 1 \pm \sqrt{b^3 - b^2 - 4b + 5}}{b-2}$

Now, observe that $x^2 + bx + 1 = (x + \dfrac{b}{2})^2 - \dfrac{b^2 - 4}{4} = \left(x + \dfrac{b}{2} + \dfrac{\sqrt{b^2 - 4}}{2}\right)\left(x + \dfrac{b}{2} - \dfrac{\sqrt{b^2 - 4}}{2}\right)$

has zeros for $b > 2$.

We can now conclude that $\exists{b} \in (1 , 2)$ for the range set to be equal, for both $f(x)$ and $\dfrac{1}{f(x)}$, otherwise $\dfrac{1}{f(x)}$ will always have a range tending to $\infty$ or $-\infty$.

Now, if $f(x) =$ maximum $\implies$ $\dfrac{1}{f(x)} = $ minimum $\implies$ $f(x)$ and $\dfrac{1}{f(x)}$ reflections about their intersection point, which happens at $f(x) = 1; f(x) = 1/f(x) \to f^2(x) = 1 \to f(x) = 1$

There are $2$ methods to find to solve for $b$ (and evaluate the range), from here:

$$f\left(\dfrac{b - 1 \pm \sqrt{b^3 - b^2 - 4b + 5}}{b-2}\right) = \dfrac{1}{f\left(\dfrac{b - 1 \mp \sqrt{b^3 - b^2 - 4b + 5}}{b-2}\right)}$$

or

Simply continue solving for $b$, in $$f(x) = 1/f(x) \dots f(x) = 1$$

But ofc the latter is much simpler.