The function f is defined by $$f(x) = \frac{x^2 + 10}{2x + 5}$$ where f has its maximum possible domain - AQA - A-Level Maths Pure - Question 10 - 2022 - Paper 3

To find the stationary points, we compute the derivative of the function:

$f'(x) = \frac{(2x + 5)(2x) - (x^2 + 10)(2)}{(2x + 5)^2}$

Setting the numerator equal to zero:

$(2x + 5)(2x) - 2(x^2 + 10) = 0$

Expanding gives:

$4x^2 + 10x - 2x^2 - 20 = 0$

$2x^2 + 10x - 20 = 0$

This simplifies to:

$x^2 + 5x - 10 = 0$

Using the quadratic formula to find the roots:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where a=1, b=5, and c=-10:

$x = \frac{-5 \pm \sqrt{(5)^2 - 4(1)(-10)}}{2(1)}$

$x = \frac{-5 \pm \sqrt{25 + 40}}{2}$

$x = \frac{-5 \pm \sqrt{65}}{2}$

This provides the x-values of the stationary points P and Q.

Considering the behavior of f(x), we note that the curve approaches its horizontal asymptote and potential extreme values. From the stationary points identified,

the range can be expressed as:

$f(x) \in \left(-\infty, \min\left(f\left(-\frac{5 + \sqrt{65}}{2}\right), f\left(-\frac{5 - \sqrt{65}}{2}\right)\right) \right) \cup \left(\max\left(f\left(-\frac{5 + \sqrt{65}}{2}\right), f\left(-\frac{5 - \sqrt{65}}{2}\right)\right), \infty\right)$