5. (i) Find, using calculus, the x coordinate of the turning point of the curve with equation $y = e^x \, \cos 4x$, \[ \frac{\pi}{4} < x < \frac{\pi}{2} \] Give your answer to 4 decimal places - Edexcel - A-Level Maths Pure - Question 5 - 2016 - Paper 3

To find the turning point of the function ( y = e^x \cos 4x ), we first need to differentiate it with respect to x:

$\frac{dy}{dx} = e^x \cos 4x - 4 e^x \sin 4x$

Setting the derivative to zero to find the critical points:

$e^x (\cos 4x - 4 \sin 4x) = 0$

Since ( e^x ) is never zero, we can simplify to:

$\cos 4x - 4 \sin 4x = 0$

This implies:

$\cos 4x = 4 \sin 4x$

To solve for 4x, we can rewrite it as:

$\tan 4x = \frac{1}{4}$

Using the inverse tangent function:

$4x = \tan^{-1}(\frac{1}{4}) + n\pi$

This gives us:

$x = \frac{1}{4} \left( \tan^{-1}(\frac{1}{4}) + n\pi \right)$

For the interval ( \frac{\pi}{4} < x < \frac{\pi}{2} ), we can calculate the value of ( x ) for n=0. The approximate value of ( \tan^{-1}(\frac{1}{4}) \approx 0.244978 ):

$x \approx \frac{1}{4} (0.244978) \approx 0.0612445 \approx 0.2463 \text{ (for n=0)}$

Calculating the approximate x value gives:

$x \approx 0.9463.$