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 confirm that P and Q are stationary points, we first find the first derivative of $f(x)$ using the quotient rule:

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

Simplifying this:

$f'(x) = \frac{2x(2x + 5) - 2(x^2 + 10)}{(2x + 5)^2}$ $= \frac{4x^2 + 10x - 2x^2 - 20}{(2x + 5)^2}$ $= \frac{2x^2 + 10x - 20}{(2x + 5)^2}$

To find stationary points, we set the numerator equal to zero:

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

Dividing through by 2 gives:

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

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{25 + 40}}{2} = \frac{-5 \pm \sqrt{65}}{2}$

Thus, the stationary points occur at:

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

These correspond to the points P and Q.

To determine the range of $f(x)$, we analyze the behavior of the function as $x$ approaches its critical points and at the endpoints of its domain. Given that:

• The function $f(x)$ is continuous and differentiable where defined.
• As $x$ approaches $-\frac{5}{2}$ from the left or right, it tends to either positive or negative infinity.

We conclude that the function reaches every value in the real number line except potentially the limit points given by evaluating $f$ at its stationary points.

In set notation, the range of $f(x)$ is:

$\text{Range of } f: \mathbb{R}$