<s>[INST] Maximum value of $\frac{3x^2+9x+17}{x^2+2x+9}$ If $x$ is real, the maximum value of $\frac{3x^2+9x+17}{x^2+2x+9}$ is? 
Is it necessary that this function will attain maximum when the denominator will be minimum? 
 [/INST] You need to solve:
$$\frac{\text{d}}{\text{d}x}\left[\frac{3x^2+9x+17}{x^2+2x+9}\right]=0\Longleftrightarrow$$
$$\frac{47+x(20-3x)}{(9+x(2+x))^2}=0\Longleftrightarrow$$
$$47+x(20-3x)=0\Longleftrightarrow$$
$$x(20-3x)=-47\Longleftrightarrow$$
$$-3x^2+20x=-47\Longleftrightarrow$$
$$3x^2-20x=47$$
If you look to it graphicly you see that the maximum is about $8.5$ so finding the solutions of the equation we get:
$$x=\frac{10+\sqrt{241}}{3}\approx8.5081$$
So:
$$\text{max}\left[\frac{3x^2+9x+17}{x^2+2x+9}\right]=\frac{35+\sqrt{241}}{16}\space\text{at}\space x=\frac{10+\sqrt{241}}{3}$$
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