A curve C has equation y = e^{2x} an x, \, x ≠ (2n + 1) \frac{\pi}{2} - Edexcel - A-Level Maths Pure - Question 3 - 2008 - Paper 6

To find the turning points of the curve, we first need to compute the derivative of the function. We have:

$\frac{dy}{dx} = e^{2x} \tan x + e^{2x} \sec^2 x.$

Setting the derivative to zero gives:

$e^{2x} \tan x + e^{2x} \sec^2 x = 0.$

We can factor out $e^{2x}$ (which is never zero) to get:

$\tan x + \sec^2 x = 0.$

Recall that $\sec^2 x = 1 + \tan^2 x$. Substituting this in, we obtain:

$\tan x + 1 + \tan^2 x = 0.$

Rearranging gives:

$\tan^2 x + \tan x + 1 = 0.$

To determine the values for which this equation holds, we can complete the square or use the quadratic formula. However, for the turning point, we need:

$\tan x + 1 = 0$

leading to:

$\tan x = -1.$

Thus, turning points occur where $\tan x = -1$.