The first part of the problem is solved by differentiating once and equating this to zero:
y = x^3 - 3x^2 +4 .dy/dx = 3x^2 - 6x .dy/dx = x(3x - 6) .
At the turning points;
x(3x - 6) = 0 (turning points occur where the gradient, dy/dx, equals zero) .
Hence, x = 0 or 2.
Inputting these x-values into the original equation yields the respective y-coridnates or the turning points. The locations are (0, 4) and (2, 0).
The nature of the turning points can be determined by finding the second derivative of the original equation:
d^2y/dx^2 = 6x - 6 .
At (0, 4), d^2y/dx^2 = -6 .At (2, 0), d^2y/dx^2 = 6 .
Therefore (0, 4) is a maximum and (2, 0) is a minimum (positive second derivative --> minimum, negative second derivative --> maximum).

(5 + 2(2^0.5))(7 - 3(2^0.5))

Related Maths A Level answers

Find ∫(x^3+x^2+6)dx.

The curve C has equation x^2 + 2xy + 3y^2 = 4. Find dy/dx.

1. (a) Express 7cosx - 24sin x in the form R cos (x + a), (b) hence what is the minimum value of this equation

Find the x and y coordinates of the turning points of the curve 'y = x^3 - 3x^2 +4'. Identify each turning point as either a maximum or a minimum.