Find the stationary points of the curve f(x) =x^3 - 6x^2 + 9x + 1

Step One: stationary/turning points are points on the curve where the gradient equals 0 (i.e. a point at which the slope changes from negative to positive, or vice versa). So we need to find the gradient of the curve by calculating f'(x):
f'(x) = 3x² - 12x + 9

Step Two: to find where the gradient equals 0, set f'(x) = 0 and solve this to find the x-coordinates of the stationary points:
From step one, we have 3x² - 12x + 9 = 0
Dividing through by 3: x² - 4x + 3 = 0
Factorising: (x - 1)(x - 3) = 0
This is satisfied for x = 1 and x = 3

Step Three: finally plug these values of x into the equation of the curve to find the f(x) values. These coordinate pairs are then the stationary points of the curve:
If x = 1, then f(x) = 1 - 6 + 9 + 1 = 5
If x = 3, then f(x) = 27 - 54 + 27 + 1 = 1
Therefore the stationary points of the curve f(x) are (1, 5) and (3, 1)