Figure 1 shows a sketch of the curve C which has equation y = e^{x^3} ext{sin } 3x, -rac{ m{ ext{Pi}}}{3} ext{ } extless x extless rac{ m{ ext{Pi}}}{3} - Edexcel - A-Level Maths Pure - Question 4 - 2012 - Paper 5

To find the x-coordinate of the turning point, we need to set the derivative of the function to zero. We begin with:

[ y = e^{x^3} \sin 3x. ]

Using the product rule:

[ \frac{dy}{dx} = e^{x^3} \cdot (3x^2 \sin 3x + 3 \cos 3x). ]

Setting the derivative to zero:

[ e^{x^3} (3x^2 \sin 3x + 3 \cos 3x) = 0. ]

Since $e^{x^3} \neq 0$, we simplify to:

[ 3x^2 \sin 3x + 3 \cos 3x = 0 ]

Dividing through by 3 gives:

[ x^2 \sin 3x + \cos 3x = 0. ]

This leads to two components:

1. ( \cos 3x = 0 )

   The solutions for this are ( 3x = \frac{\pi}{2} + n\pi ) where n is an integer. Hence: [ x = \frac{\pi}{6} + \frac{n\pi}{3}. ]

2. ( x^2 \sin 3x = 0 ) implies ( x = 0 ) or ( \sin 3x = 0 ).

   Thus, for ( \sin 3x = 0 ): [ 3x = n\pi \rightarrow x = \frac{n\pi}{3}. ]

   Next, to find the turning points for positive x:

   For the smallest positive n, when n = 1, we get ( x = \frac{\pi}{3} ).

Thus, the x-coordinate of the turning point P is ( \frac{\pi}{3}. )