The curve C has equation y = 3x^3 - 8x^2 - 3 (a) (i) Find dy dx (ii) d^2y dx^2 (b) Verify that C has a stationary point when x = 2 (c) Determine the nature of this stationary point, giving a reason for your answer. - Edexcel - A-Level Maths Pure - Question 2 - 2017 - Paper 1

To find the second derivative, we differentiate our first derivative:

$\frac{dy}{dx} = 12x^2 - 24x$

Differentiating again,

$\frac{d^2y}{dx^2} = 12 \cdot 2x^{2-1} - 24 \cdot 1$

This simplifies to:

$\frac{d^2y}{dx^2} = 24x - 24$

To verify that the curve has a stationary point at x = 2, we substitute x = 2 into the first derivative:

$\frac{dy}{dx} = 12(2)^2 - 24(2)$

Calculating this gives:

$\frac{dy}{dx} = 12 \cdot 4 - 48 = 48 - 48 = 0$

Since $\frac{dy}{dx} = 0$, this confirms there is a stationary point when x = 2.

To determine the nature of the stationary point at x = 2, we will evaluate the second derivative at this point:

From the second derivative:

$\frac{d^2y}{dx^2} = 24x - 24$

Substituting x = 2, we get:

$\frac{d^2y}{dx^2} = 24(2) - 24 = 48 - 24 = 24$

Since $\frac{d^2y}{dx^2} > 0$, this indicates that the stationary point is a local minimum.