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

${x}_{1} = 1 - \sqrt{7} , {x}_{2} = 1 + \sqrt{7}$
If you want the critical points of $k ' \left(x\right)$, you have to compute $k ' ' \left(x\right)$. An easy calculation gives us $k ' ' \left(x\right) = 3 {x}^{2} - 6 x - 18.$You only have to solve $3 {x}^{2} - 6 x - 18 = 0.$