Question

CALCULUS
Find the critical points and the local extreme values
f(x)=x^3+3x-2
f'(x)=3x^2+3
then.. f'(x)=3x^2+3=0? then 3x^2=-3 then x^2=-1 which has no points or you can do 3(x-1)(x+1)?

Answer 1

Please remember your algebra.

3x^2 + 3 = 0 or x^2 + 1 = 0 has no solution in the Real Numbers. Why would you even try? f'(x) is never zero.

Now, let's have a look at the 2nd derivative.