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? 

Note that by division $\dfrac{3x^2+9x+17}{x^2+2x+9}=3+\dfrac{3x-10}{x^2+2x+9}$, so if the first expression has a maximum (or minimum) value it will occur when $\dfrac{3x-10}{x^2+2x+9}$ has its maximum (or minimum) value. The advantage of using  $\dfrac{3x-10}{x^2+2x+9}$ is that the derivative is algebraically simpler.
