Question

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

Answer 1

Setting the derivative to zero, and solving the equation will give you the `x`'s.

Explanation:

`f'(x)=3*x^2-2*4x+1=3x^2-8x+1=0`
This is a quadratic equation, that you should be able to solve.
Then put the two `x`-values into the original function:
graph{x^3-4x^2+x+6 [-14.24, 14.23, -7.12, 7.12]}