# How do you find the coordinates of relative extrema f(x)=x^3-4x^2+x+6?

Aug 6, 2015

Setting the derivative to zero, and solving the equation will give you the $x$'s.
$f ' \left(x\right) = 3 \cdot {x}^{2} - 2 \cdot 4 x + 1 = 3 {x}^{2} - 8 x + 1 = 0$
Then put the two $x$-values into the original function: