f(x)= 3x^2+4x+1

f'(x) = 6x +4

the critical value is x values in which f'(x)=0

then 6x+4=0

==> x=-4/6=-2/3

the the critical vaue for f(x) is x=-2/3

Now f'' = 6 which is positive.

then x=-2/3 is a minimum value for the function.

f(x)= 3x^2+4x+1

f'(x) = 6x +4

the critical value is x values in which f'(x)=0

then 6x+4=0

==> x=-4/6=-2/3

the the critical vaue for f(x) is x=-2/3

Now f'' = 6 which is positive.

then x=-2/3 is a minimum value for the function.