A curve C has equation y = e^{2x} an x, ext{ where } x eq (2n + 1)\frac{oldsymbol{ ext{ ext{π}}}}{2} - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 6

To find the turning points, we need to differentiate the given equation with respect to x.

The first derivative of y is given by:

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

Setting the first derivative equal to zero to find critical points:

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

Factoring out (e^{2x}):

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

Since (e^{2x}) is never zero, we set:

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

Recall that (\sec^2 x = 1 + \tan^2 x), substituting gives:

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

Rearranging provides:

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

This factors to:

$(\tan x + 1)^2 = 0$

Thus, (\tan x + 1 = 0) implies (\tan x = -1), showing that turning points occur where (\tan x = -1).