Find the stationary points of y = 4(x^2 - 4)^3

This question challenges the student's ability to use the chain rule and to understand what a stationary point is. Having recognised that the chain rule is required, the first step is to use the substitution u = x^2-4 so that our function can be rewritten as y = 4u^3. Now, differentiating y with respect to u we find dy/du = 12u^2. Similarly, differentiating u with respect to x we find du/dx = 2x. The chain rule dictates that dy/dx = dy/du * du/dx. Inserting our expressions for dy/du and du/dx one finds that dy/dx = 24x * u^2 = 24x(x^2 - 4)^2. Stationary points occur when dy/dx = 24x(x^2 - 4)^2 = 0 so x=0 or x^2 = 4 => x = 2 or x = -2. When x = -2, y = 0; when x = 2, y = 0; when x = 0, y = -256. (Looks much prettier on a whiteboard!)