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f'(x) = 6x +4
the critical value is x values in which f'(x)=0
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.
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
Now, we'll calculate the root of the first derivative, this value being the value for the function f reaches the critical value.
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.
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