Find points at which the tangent to a given function is zero

by RoRi

September 16, 2015

Find the points at which the tangent line to the function

$f(x) = \frac{1}{3}x^3 -2x^2 + 3x + 1$

is horizontal.

First, we compute the derivative,

$f'(x) = x^2 - 4x + 3.$

Setting this equal to zero we have,

$x^2 - 4x + 3 = (x-3)(x-1) = 0 \ \ \implies \ \ x = \{ 1,3 \}.$

Stumbling Robot