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

Question

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

sburchette · Answer

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).$$0=6x^2-6x$$$$0=6(x^2-x)$$$$0=x^2-x$$$$0=x(x-1)$$$$x=0, x=1$$

anonymous · Answer

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)

sburchette · Answer

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.