can u help me out, i'm stuck with it

Question

If  f(x) = sin^4(x)cos^4(x). Then f is increasing function in the interval? 

Answer 1

I suggest using the identity 

 

sin2x = 2 sinx cosx

 

 

Then

 

sinx cosx = (sin2x)/2

 

sin⁴x cos⁴x = (1/16)sin⁴(2x)

 

So now you have

 

f(x) = (1/16)sin⁴(2x)

 

Consider the interval [0,π/4].  Note that sin(2x) is increasing, so f(x) is increasing.

From π/4 to  π/2, sin(2x) is decreasing, so f(x) is decreasing.

Continue to perform this type of analysis all the way up to 2π, and you will figure out where f(x) is increasing.  

 

Note f(x) is never negative.

 

Another approach is to find the sub-intervals of [0,2π] where f'(x) is positive and where it is negative, and that will tell you where f(x) is increasing and where f(x) is decreasing, respectively.  Try it.

Answer 2

Set the derivative of f(x) equal to zero and solve for x.

 

f'(x) = 0

 

4sin³(x)cos(x)cos⁴(x) - 4cos³(x)sin(x)sin⁴(x) = 0

 

4sin³(x)cos⁵(x) - 4sin⁵(x)cos³(x) = 0

 

4sin³(x)cos³(x)(cos²(x) - sin²(x)) = 0

 

4(1 - cos²(x))³cos³(x)(2cos²(x) - 1) = 0

 

 

Set the factors equal to zero and solve for x from each one to get your critical values.

 

1 - cos²(x) = 0         ,        cos(x) = 0         ,            2cos²(x) - 1 = 0

 

 

Then evaluate the derivative around the critical values.  If the derivative is positive around the critical values, then the function is increasing at the critical value.