A curve C has equation $$y = x^2e^x.$$ (a) Find \( \frac{dy}{dx} \) using the product rule for differentiation - Edexcel - A-Level Maths Pure - Question 5 - 2007 - Paper 5

Turning points occur where ( \frac{dy}{dx} = 0 ). Thus, we set the derivative equal to zero:

$$e^x(2x + x^2) = 0.$$

Since ( e^x \neq 0 ) for all real x, we solve:

$$2x + x^2 = 0 \Rightarrow x(2 + x) = 0.$$

This gives us:

1. ( x = 0,)
2. ( 2 + x = 0 \Rightarrow x = -2.)

Now, we substitute these x-values back into the original equation to find their corresponding y-coordinates:

1. For ( x = 0:) ( y = 0^2e^0 = 0. )
2. For ( x = -2:) ( y = (-2)^2e^{-2} = 4e^{-2} ).

Thus, the turning points of C are ( (0, 0) ) and ( (-2, 4e^{-2}) ).

To find the second derivative, we differentiate ( \frac{dy}{dx} ) again:

$$\frac{d^2y}{dx^2} = \frac{d}{dx}[e^x(2x + x^2)] \text{ using the product rule again.}$$

Letting ( u = e^x ) and ( v = 2x + x^2 ):

• ( u' = e^x, )
• ( v' = 2 + 2x. )

Thus,

$$\frac{d^2y}{dx^2} = e^x(2 + 2x) + e^x(2x + x^2) = e^x(2 + 4x + x^2).$$

To classify the turning points, we evaluate the second derivative at the turning points found in part (b).

1. For ( x = 0:) ( \frac{d^2y}{dx^2} = e^0(2 + 4(0) + 0^2) = 2 > 0, ) indicating a local minimum.

2. For ( x = -2:) ( \frac{d^2y}{dx^2} = e^{-2}(2 + 4(-2) + (-2)^2) = e^{-2}(2 - 8 + 4) = e^{-2}(-2) < 0, ) indicating a local maximum.

Thus, the nature of the turning points are:

• ( (0, 0) ) is a local minimum.
• ( (-2, 4e^{-2}) ) is a local maximum.