Find where the function f(x)=3x^4-4x^3-12x^2+5 is increasing and decreasing?

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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f(x)=3x^4-4x^3-12x^2+5

First we need to find the critical points in which the function changes behavior (increasing or decreasing)

To determine the critical points, we need to find f'(x)

f'(x)=12x^3-12x^2-24x = 12x(x^2-x-2)=12x(x-2)(x+1)

The critical points are 0, 2, -1

Then we have 4 intervals where the function change behavior

(-inf,-1), (-1,0), (0,2), (2,inf)

1. when x<-1 ==> f'(x) >0 ==> f is decreasing

2. when -1<x<0 ==> f'(x) <0 ==> f is increasing

2. when 0<x<2 ==> f'(x) < 0 ==> f is decreasing

4. when x>2 ==> f'(x)>0 ==> f is increasing

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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For this reason, we'll use the Increasing/Decreasing Test.

Of course, to use this test, we have to differentiate the function first:

f'(x)=(3x^4-4x^3-12x^2+5)

We know that the derivative of the sum is the sum of derivatives:

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

We'll factorize:

f'(x)=12x(x^2-x-1)

We'll factorize again, writing the expression

x^2-x-1=(x-2)(x+1)

f'(x)=12x(x-2)(x+1)

Now, we'll find the critical points:

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

12x=0

x=0

x-2=0

x=2

x+1=0

x=-1

And now, we'll begin to analyze the behaviour of the derivative, around these critical points:

For x<-1:

12x<0

(x-2)<0

(x+1)<0

Multiplying 12x(x-2)(x+1)=(-)*(-)*(-)<0

f'(x)<0, so f is decreasing over (-inf, -1).

For -1<x<0:

12x<0

(x-2)<0

(x+1)>0

Multiplying 12x(x-2)(x+1)=(-)*(-)*(+)>0

f'(x)>0, so f is increasing over (-1, 0).

For 0<x<2:

12x>0

(x-2)<0

(x+1)>0

Multiplying 12x(x-2)(x+1)=(+)*(-)*(+)<0

f'(x)<0, so f is decreasing over (0, 2).

For x>2:

 

12x>0

(x-2)>0

(x+1)>0

 

Multiplying 12x(x-2)(x+1)=(+)*(+)*(+)>0

f'(x)>0, so f is increasing over (2, inf).

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

f(x) = 3x^4-4x^3-12x^2+5

 To determine the point where f(x) is increasing and then decresing.

Solution:

The point wher f(x) increases on left and decrease on the right is the point where f(x) attains maximum . So  we  can   f'(c) = 0  for c and see which value of c makes f" (c) negative.

f'(x) = 0 gives: (3x^4-4x^3-12x^2+5)' = 0. Or

12x^3-12x^2-24x = 0. Or dividing by 12  we get:

x(x^2-x-2) = 0. Or

x(x-2)(x+1) = 0. Or

x=0, x =2 or x = -1

f"(x) = (12x^3-12x^2-24x)' =  36x-24x-24 = 12(3x^2-2x-2)

f(0) = 12*(-2) is - ve.

f(2) = 12(3*2^2-2*2-2) = 12*(6) is positive.

f(-1) = 12(3*(-1)^2-2(-1)-2) = 12*3 is positive.

Therefore f(0) maximum at x= 0 0, where f(x) is incresing on the left of x= 0 and f(x) decreases on the right of x= 0.

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