Question

What are the critical values of `f(x)=3x^2+2x+5`?

Answer 1

`x=-1/3,y=14/3`

Explanation:

Computing the first derivative of `f(x)` we get

`f'(x)=6x+2`
solving the equation `f'(x)=0`
we get `x=-1/3`

since `f''(x)=6>0` we get a minimum (this is global, since we have a parabola)
and we get

`f(-1/3)=3*1/9-2/3+5=(1-2+15)/3=14/3`