Question

Find the coordinates of all critical points:
f(x)=2x^3-3x^2+4

Answer 1

I'm assuming critical points are values where the derivative is equal to zero.\[f'(x)=6x^2-6x\]Since critical numbers occur where the derivative equals zero, we substitute 0 for f'(x).\[0=6x^2-6x\]\[0=6(x^2-x)\]\[0=x^2-x\]\[0=x(x-1)\]\[x=0, x=1\]

Answer 2

For the 0=x(x-1), do we assume 1 since x is understood to be 1? (Trying to figure out where 1 came from)

Answer 3

The 1 came from he (x-1). Since the two factors multiply to 0, x could equal 0 or x could equal 1. If you plug 0 or 1 into the expression x(x-1) you will get 0 either way.