Maximum value

$\begin{aligned} & =\frac{3x^2+9x+17}{3x^2+9x+7} \\ & = \frac{3x^2+9x+7+10}{3x^2+9x+7} \\ & = 1+\frac{10}{3x^2+9x+7} \end{aligned}$

Now for the above expression to get maximum value, the value of $3x^2+9x+7$ must be minimum.

The minimum value of $\underbrace{3}_{a}x^2+\underbrace{9}_{b}x+\underbrace{7}_{c}$ would be at $x=-\frac{b}{2a}=-\frac{9}{6}=-\frac{3}{2}$ .

So maximum value of $f(x)$ ,

$\begin{aligned} & = 1+\frac{10}{3x^2+9x+7} \\ & =1+\frac{10}{3\left(-\frac{3}{2}\right)^2+9\left(-\frac{3}{2}\right)+7} \\ & = 1+\frac{10}{\frac{1}{4}}=1+40=\boxed{41} \end{aligned}$

Let the expression be $f(x)$ , then we have:

$\begin{aligned} f(x) & = \frac{3x^2+9x+17}{3x^2+9x+7} \\ & = 1 + \frac{10}{3x^2+9x+7} \\ & = 1 + \frac{10}{3\left(x^2+3x+\frac{9}{4}\right) + \frac{1}{4}} \\ & = 1 + \frac{10}{\color{#3D99F6}{3\left(x +\frac{3}{2} \right)^2 + \frac{1}{4}}} \end{aligned}$

$f(x)$ is maximum, when $\color{#3D99F6}{3\left(x +\frac{3}{2} \right)^2 + \frac{1}{4}}$ is minimum. Since $\color{#3D99F6}{3\left(x +\frac{3}{2} \right)^2} \ge 0$ , $\color{#3D99F6}{3\left(x +\frac{3}{2} \right)^2 + \frac{1}{4}}$ is minimum, when $\color{#3D99F6}{3\left(x +\frac{3}{2} \right)^2}= 0$ .

$\begin{aligned} \text{Therefore, maximum } f(x) & = 1 + \frac{10}{\color{#3D99F6}{0 + \frac{1}{4}}} = 1 + 40 = \boxed{41} \end{aligned}$

Easily solvable with calculus. set f(x) = the expression. compute the derivative of the function, which is

f'(x)= (-60x-90)/(3x^2+9x+7)^2 -60x-90=0 x=-3/2 is a critical point.

You can plug in the point x=-3/2 to make sure it's continuous there, but it's pretty obvious by looking at it. Find a number smaller than -3/2 to test for f((x<-3/2)) > 0.

f(-2) = 11 >0, so that means x=-3/2 is a maximum. You can confirm by plugging in values of x < -3/2 to see that they are smaller than f(-3/2) = 41.

Here is the solution: