g(x)=3x^3+12x^2+15x
g'(x) = 9x^2 + 24x + 15
Setting g'(x) to zero, we have:
g'(x) = 0
9x^2 + 24x + 15 = 0
3x^2 + 8x + 5 = 0
(3x + 5)(x + 1) = 0
3x + 5 = 0 or x + 1 = 0
x = -5/3 or x = -1
Thus, the critical numbers are x = -5/3 and x = -1
By substitution, g'(x) > 0 for x < -5/3 and x > -1
This means g(x) is increasing for x < -5/3 and for x > -1
Similarly, g'(x) < 0 for -5/3 < x < -1
This means x is decreasing for -5/3 < x < -1