The curve C has equation $y = 6 - 3x - \frac{4}{x}$, $x \neq 0$ - Edexcel - A-Level Maths Pure - Question 9 - 2013 - Paper 6

From part (a), we've established the first turning point at $x = \sqrt{2}$. To find the value of the second turning point, we can revisit our derivative:

We also know that $y' = 0$ provides turning points. We managed to simplify down to:

$x^2 = 4 \implies x = -2$

Thus, the x-coordinate of the other turning point Q is $x = -2$.

To find the second derivative, we differentiate the first derivative again:

$y' = -3 + \frac{12}{x^2}$

Differentiating gives:

$y'' = \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\left(-3 + 12x^{-2}\right) = -24x^{-3}$

This is the simplified expression for the second derivative.

To determine the nature of the turning points, we use the second derivative test:

For $x = \sqrt{2}$:

$y'' = -24(\sqrt{2})^{-3} < 0$

This indicates a local maximum at point P.

For $x = -2$:

$y'' = -24(-2)^{-3} < 0$

This also indicates a local maximum at point Q. Therefore, both turning points P and Q are local maxima.