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

Let $y=\frac{3x^2+9x+17}{3x^2+9x+7}$$y(3x^2+9x+7)=3x^2+9x+17$$3x^2(y-1)+9x(y-1)+7y-17=0$ Since $x$ is real, the discriminant $D\ge 0$:$D=[9(y-1){]}^2-4[3(y-1)][7y-17]\ge 0$$81(y-1{)}^2-12(y-1)(7y-17)\ge 0$ Dividing by $3(y-1)$ (assuming $y\ne 1$):$27(y-1)-4(7y-17)\ge 0$$27y-27-28y+68\ge 0$$-y+41\ge 0⟹y\le 41$ The maximum value is $41$.