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

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### 4 Answers

f(x) = 6x^3 + 3x^2 + 4

First we will calculate the first derivative's zeros:

f'(x) = 18x^2 + 6x = 0

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

==> x= 0 and x = -1/3

Then we have two extreme values:

f(0) and (f(-1/3)

f(0) = 4

f(-1/3) = 6(-1/3)^3 + 3(-1/3)^2 + 4

= -6/27 + 1/3 + 4

= (-6+ 9 + 108)/27 = 111/27 = 37/9 = 4 1/9

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**

To find the extreme values of f(x) = 6x^3+3x^2+4, we find f'(x) and set f'(x) = 0 and solve for x. The solution set of x's for which f'(x) = 0 are called critical values of x. f(x) extreme for the set of critical vvalues of x.

f(x) = 6x^3+3x^2+4.

We differentiate both sides.

f'(x) = 0 gives (6x^3+3x^2+4)' = 0

3*6x^2+2*3x = 0

6x(3x+1) = 0.

x = 0 or x+1 = 0 for x = 0 or x = -1/3 are the crical values.

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

f(-1) = 6*(-1/3)^3+3(-1/3)^2+4 = -6/27+3/9+4 = 4+1/9 are the extreme values.

To find the extreme values of the functions f(x)= 6x^3+ 3x^2 +4, we first differentiate the function.

Here we will use a general relation which is valid whenever we are trying to differentiate a function. It states that if f(x) = a*x^n, then f’(x) is equal to a*n*x^(n-1) .

Therefore f'(x) = 6*3*x^2+3*2*x = 18x^2 + 6x

Now equate this to zero,

18x^2 + 6x = 0

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

Or x can be 0 or -1 / 3.

Now differentiate f'(x) again, we get 36x + 6

For x=0, 36x + 6 = 6 is positive, therefore at x=0 we have a minima.

For x= -1/3, 36x + 6 = 36*(-1/3) +6= -12+6 =-6 . As this is negative at x= -1/3 we have a maxima.

Therefore the maxima of the function is at x=-1/3 and the minima is at x=0.