`f(x) = 3x^4 - 4x^3 - 12x^2 + 1, [-2, 3]` Find the absolute maximum and minimum values of f on the given interval

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Given: `f(x)=3x^4-4x^3-12x^2+1,[-2,3].`

Find the critical numbers by setting the first derivative equal to zero and solving for the x value(s).

`f'(x)=12x^3-12x^2-24x=0`

`12x(x^2-x-2)=0`

`12x(x-2)(x+1)=0`

`x=0,x=2,x=-1` 

The critical numbers are x=0,x=2, and x=-1. Plug in the critical numbers and the endpoints of the interval [-2,3] into the original f(x) function.

f(-2)=33

f(-1)=-4

f(0)=1

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Given: `f(x)=3x^4-4x^3-12x^2+1,[-2,3].`

Find the critical numbers by setting the first derivative equal to zero and solving for the x value(s).

`f'(x)=12x^3-12x^2-24x=0`

`12x(x^2-x-2)=0`

`12x(x-2)(x+1)=0`

`x=0,x=2,x=-1` 

The critical numbers are x=0,x=2, and x=-1. Plug in the critical numbers and the endpoints of the interval [-2,3] into the original f(x) function.

f(-2)=33

f(-1)=-4

f(0)=1

f(2)=-31

f(3)=28

Examine the f(x) values to determine the absolute maximum and absolute minimum.

The absolute maximum value occurs at the point (-2, 33).

The absolute minimum value occurs at the point (2, -31).

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`f(x)=3x^4-4x^3-12x^2+1`

differentiating,

`f'(x)=12x^3-12x^2-24x`

`f'(x)=12x(x^2-x-2)`

Now to find the absolute extrema of the function , that is continuous on a closed interval, we have to find the critical numbers that are in the interval and evaluate the function at the endpoints and at the critical numbers.

Now to find the critical numbers, solve for x for f'(x)=0.

`12x(x^2-x-2)=0`

`12x(x+1)(x-2)=0` 

`x=0 , x=-1 , x=2`

Now let us evaluate the function at the critical numbers and at the end points of the interval (-2,3).

`f(0)=1`  

`f(-1)=3(-1)^4-4(-1)^3-12(-1)^2+1=-4`

`f(2)=3(2)^4-4(2)^3-12(2)^2+1=-31`

`f(-2)=3(-2)^4-4(-2)^3-12(-2)^2+1=33`

`f(3)=3(3)^4-4(3)^3-12(3)^2+1=28`

The function has a absolute minimum=-31 at x=2

Function has absolute maximum=33 at x= -2

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