What is the extreme values of the function f(x)= 6x^3+ 3x^2+ 4

To calculate the local extreme values of a function, we'll
have to do the first derivative test.

To do this test,
we'll have to differentiate the function:

 f'(x)= (6x^3+
3x^2+ 4)'

If you are familiar with the derivative of the
functions, we'll solve the problem in this way:

 f'(x)=
(6x^3)'+ (3x^2)'+ (4)'

 f'(x)= 18x^2 +
6x

The values of x for the first derivative is cancelling
are the extreme values for the function f(x).

Let's
calculate the derivative zeroes:

18x^2 + 6x =
0

We'll factorize by
6x:

(6x)(3x + 1) = 0

We'll set
each factor as zero:

6x =
0

We'll divide by 6:

x =
0

3x + 1 = 0

We'll subtract 1
both sides:

3x = -1

We'll
divide by 3:

x = -1/3

The
extreme values of the function are:

f(0) = 6*0^3+ 3*0^2+
4

f(0) =
4

f(-1/3) = 6*(-1/3)^3+ 3*(-1/3)^2+
4

 f(-1/3) = -6/27 + 1/3 +
4

 f(-1/3) = -2/9 + 1/3 +
4

 f(-1/3) = (-2 + 3 +
36)/9

 f(-1/3) =
37/9