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

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).