Find condition on $a$ and $b$ such that $h(x) = \frac{x^2 - x +1}{x^2 + ax+(1-b^2)}$ is bounded

Question

$$ h(x) = \frac{x^2 - x +1}{x^2 + ax+(1-b^2)}$$

So, basically, I have to find the values of parameters 'a' and 'b' such that this function has both upper and lower limit.

What I've done:

$$ h(x) = 1 - \frac{ (1+a)x +b^2}{x^2 + ax+(1-b^2)}$$

let,

$$ g(x) = \frac{ (1+a)x +b^2}{x^2 + ax+(1-b^2)}$$

$h$ is optimized when $g$ is optimized

since 'g' is a quotient, Quotient functions like $ f=\frac{u(x)}{v(x)}$ are optimized for all x satisfying the criterion when $ 0 = u'(x) v(x) - v'(x) u(x)$ , so, for optimum of 'h',

$$ 0= (1+a) [ x^2 + ax + (1-b^2) ] - [ 2x +a ] ( (1+a) x +b^2)$$

Expanding this out, I got,

$$ 0 = x^2 [ 3a+3 ] + x [ 2a^2 -2b^2 + 2a] + [1-b^2 + a - 2ab^2]$$

Now, I'm not sure how to figure out the rest, if I were to guess, I need to see how the functions behaves if I plug in the roots like if it blows up or not. If it blows up then I have to set discriminant less than 0, but for checking if it blows up, I'll have to solve for roots... which seems..painful? is there a better way or am I missing something?

Answer 1

The nominator $$x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0,$$ $$\lim_{x\rightarrow+\infty}h(x)=\lim_{x\rightarrow-\infty}h(x)=1.$$ Thus, since $h$ is a continuous function, $h$ would be bounded iff the denominator not equal to $0$ for any $x$, which gives $$a^2-4(1-b^2)<0$$

Answer 2

Hint :

If the denominator is always non-zero, then $h$ is defined and continuous on $\mathbb{R}$, and tends to $1$ in $-\infty$ and in $+\infty$.

If the denominator can be zero, then $h$ is unbounded (... really ?).