I had this problem in an exam I recently appeared for:
Find the range of
$$y =\frac{x^2+2x+4}{2x^2+4x+9}$$
By randomly assuming the value of $x$, I got the lower range of this expression as $3/7$. But for upper limit, I ran short of time to compute the value of it and hence couldn't solve this question.
Now, I do know that one way to solve this expression to get its range is to assume the whole expression as equals to K, get a quadratic in K, and find the maximum/minimum value of K which will in turn be the range of that expression. I was short on time so avoided this long winded method.
Another guy I met outside the exam center, told me he used an approach of $x$ tending to infinity in both cases and got the maximum value of this expression as $1/2$. But before I could ask him to explain more on this method, he had to leave for his work.
So, will someone please throw some light on this method of $x$ tending to infinity to get range, and how it works. And if there exists any other efficient, and quicker method to find range of a function defined in the form of a ( quadratic / quadratic ).
Best Answer
The question can be easily solved by this technique:
As $\displaystyle y = \frac {x^2 + 2x + 4}{2x^2 + 4x + 9} \implies 2y = \frac {2x^2 + 4x + 9 - 1}{2x^2 + 4x + 9}$.
Thus, $\displaystyle 2y = 1-\frac {1}{2(x + 1)^2 + 7} $
Squares can never be less than zero so the minimum value of the function : $\displaystyle 2(x + 1)^2 + 7 $ would be $7$ , or Maximum value of $\displaystyle \frac {1}{2(x + 1)^2 + 7} $ is $\displaystyle \frac {1}{7} $.
This tells that minimum value of $y $ will be $\displaystyle \frac{3}{7}$.
And so on.. check for $x \rightarrow \infty$.
From here you can easily tell the maximum and minimum values : $\displaystyle y \in \left [ \frac {3}{7}, \frac {1}{2} \right ) $