Find the maximum or minimum for f(x)=3x^2+x+1

Answer 1

using Calculus ...
f'(x) = 6x + 1

6x+1 = 0
x = -1/6
f(-1/6) = 3(1/36) - 1/6 + 1 = 11/12

since the parabola opens upwards,
the minimum value is 11/12