If $x$ is real, compute the maximum integer value of
\[\frac{3x^2 + 9x + 17}{3x^2 + 9x + 7}.\]
So we have a function $f(x)$ and we want to find the maximum integer value of $f(x)$. Let's first simplify $f(x)$. We have that
\[f(x) = \frac{3x^2 + 9x + 17}{3x^2 + 9x + 7} = 1 + \frac{10}{3x^2 + 9x + 7}.\] So we want to find the maximum integer value of $1 + \frac{10}{3x^2 + 9x + 7}$. Let's call $g(x) = \frac{10}{3x^2 + 9x + 7}$. So we want to find the maximum integer value of $1 + g(x)$. Right. Let's find the domain of $g(x)$. We can find where the denominator is 0 by solving the equation $3x^2 + 9x + 7 = 0$. That equation has no real solutions because the discriminant is negative. That means the maximum value of $g(x)$ occurs either at a critical point or at an endpoint of the domain of $g(x)$. We can find the critical points of $g(x)$ by solving the equation $g'(x) = 0$. $g'(x)$ is the derivative of $g(x)$. So $g'(x) = 0$ when $x = -\frac{3}{2}$. That is the only critical point. The domain of $g(x)$ is the set of all real numbers, so the endpoints of the domain are $-\infty$ and $+\infty$. We know that $\lim_{x \to -\infty} g(x) = 0$ and $\lim_{x \to \infty} g(x) = 0$. So the maximum value of $g(x)$ must occur at $x = -\frac{3}{2}$. $g(-\frac{3}{2}) = \frac{10}{\frac{1}{4}} = 40$. So the maximum value of $g(x)$ is 40, which means the maximum integer value of $1 + g(x)$ is $1 + 40 = 41$.

