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? 
 A: The denominator is
$$
\underbrace{x^2+2x+9 = (x^2+2x+1)+8}_\text{completing the square} = (x+1)^2 + 8 = (\text{a square}) + 8
$$
and that is always positive, never $0$.  Therefore this function is everywhere continuous.  As $x\to\pm\infty$, the function approaches $3$.  Consequently, it must have a global maximum value unless it is always less than $3$.
If there is is a global maximum value, then, since the function is everywhere differentiable, the maximum must occur at a point where the derivative is $0$.  If there are only finitely many points where the derivative is $0$, then find the value of the function at each of them.  If the value is more than $3$ at one such point, then there is a maximum value, and you just have to pick the biggest value among those finitely many.
The derivative is a fraction, and a fraction is $0$ only if the numerator is $0$.
A: 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.
A: As the denominator is positive, we only need to find $M$ that allows equality in:
$$3x^2+9x+17 \le M(x^2+2x+9)$$
$$\iff (M-3)x^2+(2M-9)x+(9M-17) \ge 0$$
For this to hold for large $|x|$, we need $M> 3$. Further, for equality, the LHS must touch the $X$ axis, so we need 
$$\triangle = 0 \implies (2M-9)^2=4(M-3)(9M-17) \implies M = \frac{35+\sqrt{241}}{16} \approx 3.158$$
A: 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:
$$\text{max}\left[\frac{3x^2+9x+17}{x^2+2x+9}\right]=\frac{35+\sqrt{241}}{16}\space\text{at}\space x=\frac{10+\sqrt{241}}{3}$$
