Question

How do you find the critical points for `y = 2x^3 + 3x^2 - 12x + 4`?

Answer 1

The critical points are `x=-2` and `x=1`.

You evaluate the first derivative and find its zeroes:

`y'=2*3x^2+3*2x-12*1+0=6x^2+6x-12`
`y'=6x^2+6x-12`
And now we can use the quadratic formula
`Delta=6^2-4*6*(-12)=324`
`sqrt(Delta)=18`
`x_1=(-6-18)/(2*6)=(-24)/12=-2`
`x_2=(-6+18)/(2*6)=12/12=1`