The critical value for a function is that point c, which belongs to the domain, for f'(c)=0 or f'(c) does not exist.

To calculate the critical value for a function, we have to calculate the first derivative of the function, which, in this case, is:

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

f'(x)= 3*2*x + 4*1 + 0

f'(x)= 6x+4

Now, we'll calculate the root of the first derivative, this value being the value for the function f reaches the critical value.

6x+4=0

x=-4/6

x=-2/3

**The critical value of the function is: x=-2/3.**

The critical values of f(x) = 3x^2+4x+1 is got by equating f'(x) to 0 and f(x) to zero.Or the critical values are those values of x for which the curve crosses the x axis or attains its extrme values.

f(x) = 0 gives (3x+1)(x+1) = 0 Or 3x+1 = 0 and x+1 = 0 Or

x=-1/3 and x = -1.

f'(x) = 0 gives (3x^2+4x+1)' = 0. Or 6x+4 = 0 So x =-4/6.

So the critical values are x=-1, x -4/6 and x = -1/3.