Asked by missy
Find the maximum or minimum for f(x)=3x^2+x+1
Answers
Answered by
Reiny
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
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
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