Let
\[f(x) = \frac{9x^3+4x^2+11x+7}{x^2+bx+1}.\]
Find the greatest integer value of $b$ for which the domain of $f(x)$ is the set of all real numbers.
1 answer
The denominator of $f(x)$ can be factored as \[x^2+bx+1=(x^2+1)-2\left(-\frac{b}{2}\right)x=x^2+1-2\left(-\frac{b}{2}\right)x=x^2-2\left(-\frac{b}{2}\right)x+1.\]We can then complete the square in the quadratic $x^2-2\left(-\frac{b}{2}\right)x+1$ as \[\left(x-\left(-\frac{b}{2}\right)\right)^2+1-\left(-\frac{b}{2}\right)^2=\left(x-\left(-\frac{b}{2}\right)\right)^2+\frac{b^2}{4}+1.\]Since $\left(x-\left(-\frac{b}{2}\right)\right)^2$ is always at least zero, the smallest possible value of $x^2-2\left(-\frac{b}{2}\right)x+1$ is $\frac{b^2}{4}+1$; but since this expression is always at least $1$, the smallest possible value of $x^2+bx+1$ is $\frac{b^2}{4}+1+1=\frac{b^2}{4}+2$. Therefore, $b$ must be chosen so that $\frac{b^2}{4}+2$ is positive. This occurs when $b^2+8>0$, so we have $b^2>-8$. The largest integer satisfying this inequality is $\boxed{2}$.
1 answer