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