A curve C has equation $y = x^2 e^x$ - Edexcel - A-Level Maths Pure - Question 5 - 2007 - Paper 5

Turning points occur where $\frac{dy}{dx} = 0$. Setting our derived function to zero gives:

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

Since $e^x$ is never zero, we solve:

• $x = 0$
• $x + 2 = 0 \implies x = -2$

Calculating the corresponding $y$ values:

• For $x = 0$: $y = 0^2 e^0 = 0$
• For $x = -2$: $y = (-2)^2 e^{-2} = 4e^{-2}$

Thus, the turning points are:

• $(0, 0)$
• $(-2, 4e^{-2})$

To find the second derivative, we differentiate $\frac{dy}{dx} = e^x x (x + 2)$ again: Using the product rule again:

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

Calculating:

• $\frac{d}{dx}(e^x) = e^x$
• Using the product rule on $x(x + 2)$, we get:
  • $\frac{d}{dx}(x^2 + 2x) = 2x + 2$

So,

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

To determine the nature, we will evaluate the second derivative at the turning points:

• For $x = 0$: $\frac{d^2y}{dx^2}\bigg|_{x=0} = e^0 (0^2 + 4(0) + 2) = 2 > 0$ (indicating a local minimum)

• For $x = -2$: $\frac{d^2y}{dx^2}\bigg|_{x=-2} = e^{-2} ((-2)^2 + 4(-2) + 2) = e^{-2} (4 - 8 + 2) = e^{-2} (-2) < 0$ (indicating a local maximum)

Thus, the nature of the turning points is:

• At $(0, 0)$: Local minimum
• At $(-2, 4e^{-2})$: Local maximum