find the horizontal tangents of the curve and show how.

y=x^3-2x^2+x+1

1 answer

y = x^3 - 2x^2 + x + 1

Horizontal tangents have slope = 0

y' = 3x^2 - 4x + 1 = (3x-1)(x-1)

so, at x = 1/3 or 1 the slope is zero.

So, the lines are

y = 13/9

y = 1