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 ) $