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 the function ( y ) is increasing, we need to find when ( \frac{dy}{dx} > 0 ).

From our previous calculation, we have:

$\frac{dy}{dx} = \frac{-8x^2 - 24x + 14}{(4x^2 + 7)^2}$

Since the denominator ( (4x^2 + 7)^2 ) is always positive, we focus on the numerator:

( -8x^2 - 24x + 14 > 0 ).

Rearranging gives:

( 8x^2 + 24x - 14 < 0 ), which simplifies to ( 4x^2 + 12x - 7 < 0 ).

Now we can find the roots of the quadratic equation ( 4x^2 + 12x - 7 = 0 ) using the quadratic formula:

$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} = \frac{{-12 \pm \sqrt{{(12)^2 - 4 \cdot 4 \cdot (-7)}}}}{{2 \cdot 4}} = \frac{{-12 \pm \sqrt{{144 + 112}}}}{8} = \frac{{-12 \pm \sqrt{256}}}{8} = \frac{{-12 \pm 16}}{8}$

Calculating the roots gives us:

• ( x = \frac{4}{8} = \frac{1}{2} )
• ( x = \frac{-28}{8} = -\frac{7}{2} )

This tells us that the quadratic ( 4x^2 + 12x - 7 ) will be negative between its roots, hence:

( -\frac{7}{2} < x < \frac{1}{2} )

Thus, ( y ) is increasing when ( 4x^2 + 12x - 7 < 0 ).