Find the coordinates of any stationary points of the curve y(x)=x^3-3x^2+3x+2

A stationary point is a point where the gradient of a curve is 0. The derivative of a curve gives us a function for the gradient at every point on the curve. So we have that dy(x)/dx=0 if and only if the curve is stationary at that point. The derivative of ax^n is nax^(n-1), so we find that dy/dx=3(x^2)-32x^1+3. There are only 3 terms as the derivative of 2 is 0, since 2 can be written as 2x^0 (any number to the power of 0 is 1), and 20*x^(-1)=0.We want to find the stationary points, so we are interested in where dy/dx=0. 0=3x^2-6x+3.This is just a quadratic equation, and can be solved as normal. To simplify, divide by 3,0=x^2-2x+1.Then we look for factorizations. We look for 2 numbers that add to make -2 and multiply to make 1. -1 and -1 satify this, so we can rewrite the quadratic as0=(x-1)(x-1)=(x-1)^2. this tells us when x=1, our equation is satisfied and the gradient is 0. Now we need to find out what y is when x =1. Putting x=1 into y(x) givesy(1)=(1)^3-3(1)^2+3(1)+2=1-3+3+2=3. So the coordinate of the only stationary point is (1,3)