F(x) = 2x^3 + x^2 - 11x
Don't know if you have had Calculus.....but we can find these quickly
Take the derivative and set to 0
F' (x) = 6x^2 + 2x - 11 = 0
The solutions to this are x = -1.53 and x = 1.2
The second derivative is
12x + 2
Plugging the first x value into this results in a negative.....so we have a relative max at x = -1.53
And plugging the second value into this results in a positive.....so we have a relative min at x =1.2
Putting (-1.53) into the original function results in y = 12.01
Putiing (1.2) into the original function results in y = -8.3
So.....the rel max is (-1.53, 12.01) and a rel min at (1.2, -8.3)
So....the second answer is correct