If x is real, find the maximum value of $\frac{3x^2+9x+17}{3x^2+9x+7}$

Let $y=\frac{3x^2+9x+17}{3x^2+9x+7}=1+\frac{10}{3x^2+9x+7}$

$y\le 1+10.4=41$ Hence, the maximum value of y is 41.

where, minimum value of 3x² + 9x + 7 is $\frac{1}{4}$

Thus, the maximum value of $\frac{1}{3x^2+9x+7}$ is