Maximum value of f(x)=16*cos(2x) from [0,1]?

Question

What is the maximum value on the given interval.

Correct me where I'm wrong:

Endpoints:
f(0)=16
f(1)= -6.658

f'(x) = -32sin(2x)
-32sin(2x)=0

OK, this is where I'm spacing out.
Am I considering all possible values of x between 0 and 1 in the last equation?

Answer 1

Yes you are considering all values between 0 and 1 for the derivative.

df/dx = -32sin(2x) is correct
-32sin(2x) = 0 is also correct

Solve it:
sin(2x) = 0 ... divide each side by -32
sin is 0 at 0 so 2x = 0 and x = 0

This is a trivial case since the answer can be seen by inspection but the method of taking the derivative is of course general and can be used in much more demanding situations.

Answer 2

Your f(0) and f(1) are correct. 2x is in radian.
-32sin(2x)=0 implies sin(2x)=0
since sin(0)=0, 2x=0 or x=0
So, in the interval [0,1], 0 is the value for which f(x) is maximum. I hope I'm right. Wait till you see other answers as well.

Answer 3

Cos(2x) is maxed at 2x= 0 or x = 0. => max value = 16

Your mistake is multiplying 16 * 2 . 2 affects the frequency and period not the amplitude