Find the stationary point(s) of the curve: y = 3x^4 - 8x^3 - 3.

Firstly. Recognise which method you should use to approach this question. In this case, you can find the stationary point of a curve where its gradient is 0 i.e. at a point where the gradient changes from positive to negative or vice versa. This can be done by differentiating y (finding f'(x)) and equating to 0 (f'(x)=0) to then solve and find the x values. Let's take it step by step.
Secondly. Differentiate curve y.f(x) = 3x^4 - 8x^3 - 3f'(x) = 12x^3 - 24x^2
Thirdly. Equate to 0 and factorise the derivative (f'(x)) to make it easier to solve.12x^3 - 24x^2 = 012x^2(x - 2) = 0Treating both terms separately:12x^2 = 0 ----> x = 0x - 2 = 0 ----> x = 2
Finally. Conclude that the stationary points for curve y are found at x=0 and x=2.