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.
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 $\boxed{2}$.
To determine the domain of a rational function, we need to consider the values of $x$ for which the denominator is equal to zero. These are the values that would make the denominator undefined, as division by zero is undefined.
In this case, the denominator is $x^2+bx+1$. For the domain to include all real numbers, we need to find the values of $b$ for which the quadratic equation $x^2+bx+1=0$ has no real solutions.
We can use the discriminant of the quadratic equation to determine the nature of its solutions. For a quadratic equation $ax^2+bx+c=0$, the discriminant is given by $\Delta = b^2-4ac$.
In this case, $a=1$, $b=b$, and $c=1$, so the discriminant becomes $\Delta = b^2-4(1)(1) = b^2-4$.
For the quadratic equation to have no real solutions, the discriminant must be negative, i.e., $\Delta < 0$.
Therefore, we have $b^2-4 < 0$, which simplifies to $b^2 < 4$. Taking the square root of both sides (remembering to consider both the positive and negative square root), we get $-2 < b < 2$.
Since we are looking for the greatest integer value of $b$, the largest integer that satisfies $-2 < b < 2$ is $b=\boxed{1}$.
To determine the domain of the function $f(x)$, we need to find the values of $x$ for which the denominator $x^2+bx+1$ is not equal to zero. This is because division by zero is undefined.
We can find the values of $x$ that satisfy $x^2+bx+1=0$ by using the quadratic formula. The quadratic formula states that for a quadratic equation of the form $ax^2+bx+c=0$, the solutions for $x$ are given by:
\[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}.\]
In our case, we have $a=1$, $b=b$, and $c=1$. Substituting these values into the quadratic formula, we have:
\[x = \frac{-b \pm \sqrt{b^2-4(1)(1)}}{2(1)}.\]
Simplifying further, we have:
\[x = \frac{-b \pm \sqrt{b^2-4}}{2}.\]
For the function $f(x)$ to have a defined domain, we need to find the values of $b$ that result in real solutions for $x$. This means that the discriminant $b^2-4$ must be nonnegative, or $b^2 \geq 4$.
The greatest integer value of $b$ for which the domain of $f(x)$ is the set of all real numbers is the greatest integer less than or equal to the square root of 4. Since $\sqrt{4}=2$, the greatest integer value of $b$ is $\boxed{2}$.