A curve has equation $y = \frac{2x + 3}{4x^2 + 7}$ 9 (a) (i) Find $rac{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 the curve is increasing, we need to analyze when the derivative rac{dy}{dx} > 0:

$\frac{-8x^2 - 24x + 14}{(4x^2 + 7)^2} > 0$

Since $(4x^2 + 7)^2 > 0$ for all $x$, we focus on the numerator:

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

Multiplying through by -1 (which reverses the inequality):

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

Factoring or using the quadratic formula to solve for the roots:

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

where $a = 8$, $b = 24$, and $c = -14$. After calculation, we find the roots are at:

$x = \frac{-24 \pm \sqrt{576 + 448}}{16} = \frac{-24 \pm \sqrt{1024}}{16} = \frac{-24 \\pm 32}{16}$

This gives us:

• $x = 0.5$ and $x = -3.5$

Next, we need to test intervals based on the roots to determine where the inequality holds:

• For $x < -3.5$: Choose $x = -4$;
• For $-3.5 < x < 0.5$: Choose $x = 0$;
• For $x > 0.5$: Choose $x = 1$.

Evaluating these intervals shows that the inequality holds for:

$-3.5 < x < 0.5$

Thus:

The condition $4x^2 + 12x - 7 < 0$ indicates that $y$ is increasing in the interval $(-3.5, 0.5)$.