1. ## question about Mean Value Theorem

Hello Forumites,
I want to show that f(x)= $cos^2(x)+cos^2(π/3+x)-cos(x)*cos(π/3+x)$ is a constant function. What is it's value?
Solution

Now,to prove f(x) is a constant function,f'(x)=0 but f'(x)=-sin(2x)-sin(2*(3x+π)/3)+sin((6x+π)/3).Now what would be a interval I in which f(b)= f(a)?

2. ## Re: question about Mean Value Theorem

Originally Posted by Vinod
Hello Forumites,
I want to show that f(x)= $cos^2(x)+cos^2(π/3+x)-cos(x)*cos(π/3+x)$ is a constant function. What is it's value?
Solution

Now,to prove f(x) is a constant function,f'(x)=0 but f'(x)=-sin(2x)-sin(2*(3x+π)/3)+sin((6x+π)/3).Now what would be a interval I in which f(b)= f(a)?
If you just work with f(x) and make use of the expansion of cos(a+b) = cos(a) cos(b)-sin(a) sin(b) and the Pythagorean identity, you'll find that f(x) = 3/4 which is constant.
I can't see any need to find f'(x).

3. ## Re: question about Mean Value Theorem

What, exactly, are you trying to prove? If f(x) is constant, it has the same value everywhere and its derivative is 0 everywhere. There is NO particular interval to work with nor is the "mean value theorem" particularly relevant here.