Are you smarter than me?

$\frac{x + 2}{2x^{2} + 3(x + 2)}$

$= \frac{x+2}{2(x^{2}-4) + 3(x + 2) + 8}$

$= \frac{1}{2(x - 2) + 3 + \frac{8}{x + 2}}$

$= \frac{1}{2(x + 2) + \frac{8}{x + 2} - 5}$

we will consider maxima for positive real values (can you answer why?)

$f(x)_{min} = \frac{2(x + 2) + \frac{2^{3}}{x + 2}}{2} =\sqrt{2^{4}} = 8$

Thus maximum = $\frac{1}{8 - 5} = \frac{1}{3}$

Let $y = \frac{x+2}{2x^2 + 3x + 6}$ .

Now,

$2yx^2 + 3yx + 6y = x + 2$ . $\Rightarrow 2yx^2 + (3y -1)x + 6y - 2 = 0$

Let this be a quadratic equation in $x$ . We have to know where this equation has real roots. So we will use the discriminant.

$(3y-1)^2 - 4(2y)(6y -2) \ge 0$ .

Solving this inequality, we will get $y \in \left[\frac{1}{13},\frac{1}{3}\right]$ . So our max is $\frac{1}{3}$ .

$f(x)= \frac{x+2}{2x^2+3x+6} \\ f' (x)= \frac{ 2x^2+3x+6-(x+2)(4x+3)}{...} = 0 \\2x^2-4x^2 +3x - 11x =0 \\ x=0~~ or ~~x=-4 \\x=0 ~~give~ Max.~~ Max =f(0)=1/3\\ \boxed{1/3}$

find the critical pts. then you got the answer, maxima minima if im right :D.....

then zero is the value of x

common sense and logic are needed heheheh

f(x)=(x+2)/(2x2+3x+6) we study f we find that f(0) is the max value of f(x) and f(0)= 1/3

Find first derivative and equate =0 ; to find values of x, they are 0 and -4. And check the sign of second derivative for values of x. If second derivative is -ve at particular value of x (here it is 0), put this value at place of x to find max. value.