Let f(x)=3x^2-2x+1. At what point (x) is the tangent to f(x) horizontal?

Question

How do I solve this? 

Answer 1

To add to Brian's answer, this point is also a local extrema.  This means that the point can be either a local maximum or local minimum.  To determine which one it is, we evaluate the derivative of f(x) when x=0 and x=1.

If the derivative is positive at x=0   and   negative at x=1, then the point is a local maximum.
If the derivative is negative at x=0   and  positive at x=1, then the point is a local minimum.

Answer 2

f'(x) = 6x - 2

When the tangent is horizontal, f'(x) = 0

Therefore, x = 1/3.

Sub x = 1/3 to f(x), we get f(x) = 2/3

The coordinates of the point (x) = (1/3, 2/3)