Question

How do you find the critical points of `k'(x)=x^3-3x^2-18x+40`?

Answer 1

`x_1 = 1- sqrt(7), x_2 = 1+ sqrt(7)`

Explanation:

If you want the critical points of `k'(x)`, you have to compute `k''(x)`. An easy calculation gives us `k''(x) = 3x^2 - 6x - 18.`You only have to solve `3x^2 - 6x - 18=0.`