Expert: Paul Klarreich - 2/5/2007

Question
My problem is f(x)= xcosx  {0,pi/2}

I'm trying to find what c is equal to.  If I take the derivative I get cosx-xsinx. So, If I make that
f'(c)= cosc + c sinc=0  How do I proceed from here to solve for  c?  Thanks.

Answer
Questioner:   Colleen
Category:  Calculus
 
Subject:  Calculus - Rolle's Theorem
Question:  My problem is f(x)= xcosx  {0,pi/2}

I'm trying to find what c is equal to.  If I take the derivative I get cosx-xsinx. So, If I make that
f'(c)= cosc + c sinc=0  How do I proceed from here to solve for  c?  Thanks.
==========================================================
Hi, Colleen,

Now you have:

cos c - c sin c = 0

to solve.  That's equivalent to  

cot c = c

Unfortunately, there isn't any exact method of solving this.  (At least I can't think of one.)  But you will be able to argue that:

f'(0) = cos 0 - 0 sin 0 = 1
f'(pi/2) = cos(pi/2) - pi/2 sin(pi/2)
        = 0 - (pi/2)(1) = - pi/2
So f'(x) goes from positive to  negative in the interval and since it is continuous, must have a zero, which is in the open interval (0,pi/2).