A curve has equation $y = \frac{2x + 3}{4x^2 + 7}$ 9 (a) (i) Find $\frac{dy}{dx}$ 9 (a) (ii) Hence show that $y$ is increasing when $4x^2 + 12x - 7 < 0$ - AQA - A-Level Maths Pure - Question 9 - 2017 - Paper 1

To determine when $y$ is increasing, we need to find when $\frac{dy}{dx} > 0$. Since the denominator $(4x^2 + 7)^2$ is always positive for all real $x$, we need to focus on the numerator:

$-8x^2 - 24x + 14 > 0$

Factoring or using the quadratic formula may help us here. We rearrange the expression:

$8x^2 + 24x - 14 < 0$

To solve the inequality, we first find the roots of the equation:

$8x^2 + 24x - 14 = 0$

Using the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 8$, $b = 24$, and $c = -14$:

$x = \frac{-24 \pm \sqrt{24^2 - 4 \cdot 8 \cdot (-14)}}{2 \cdot 8}$
$= \frac{-24 \pm \sqrt{576 + 448}}{16}$
$= \frac{-24 \pm \sqrt{1024}}{16}$
$= \frac{-24 \pm 32}{16}$

The solutions are: $x = \frac{8}{16} = \frac{1}{2}$ and $x = \frac{-56}{16} = -3.5$

The intervals to test are $(-\infty, -3.5)$, $(-3.5, 0.5)$, and $(0.5, \infty)$. Testing a point in each interval will show where the quadratic is less than 0. After performing the tests, we conclude that:

$\frac{dy}{dx}$ is positive when $-3.5 < x < 0.5$, indicating that $y$ is increasing in this region. Therefore, this implies that the condition $4x^2 + 12x - 7 < 0$ holds for: $x \in \left(-\frac{7}{2}, \frac{1}{2}\right)$