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Differential Calculus

I asked a question earlier about open and closed intervals, so why did my teacher say this was wrong?
 Determine any values of c in the interval [0,2pi] for which f'(x) = 0.
f(x) = cos x
My answer was:
f'(x) = -sin x
set equal to 0
-sin x = 0
x = 0, pi, 2pi
she said
0/7 points: 0 and 2pi are not acceptable answers


Did she say why 0 and 2pi are not acceptable?  It appears to me that this is the correct solution.
Tom F.  - I agree that the answer appear to be correct.  The form [0,2pi] is written for a closed interval which would include 0 and 2pi.  If it was written as an open interval (0,2pi) or ]0,2pi[, then 0 and 2pi would not be included. 
Hi Adam,
Looks like we are all confused.  It certainly seems to me that 0 and 2pi should be part of the solution.
Would love to know why your teacher is saying  they are not?
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2 Answers

It appears to me that this is a problem on the mean value theorem.  The theorem stipulates that there is point in the OPEN interval where the slope is equal to the average rate of change over the CLOSED interval [a,b].  The average rate of change is f(a)-f(b)/(a-b).  If f(a)=f(b) then there is at least one point in the OPEN interval where the slope is 0.  Adam probably didn't describe the whole question as asked by his teacher.


She asked if the function can have Rolle's Theorem applied and if it can, find any values of c where f'(c) = 0
You confirmed my suspicion that this problem is related to mean value theorem or a special case of it in Rolle's theorem.  Both theorems stipulate that there is a point c on the open (emphasis on open) interval where slope equals the average rate of change on the closed interval [a,b]
That makes sense. So in other words it doesn't matter that it is closed, when you use Rolle's Theorem your always looking on the open interval. Thanks so much Imtiazur!
If the problem says to find a point c as stipulated in Rolle's theorem, then you must specify only those points that are on the open interval.  Your teacher will take points from you if you show x values in the closed (but not on the open interval) or outside the interval.
Hi Adam, it might have something to do with the two points being the boundary points and that if you think of derivatives in terms slops then you will have infinite slops which would man that the slope at those points is undefined.
Hope this helps.


Not sure how you concluded that the boundary points will have infinite slopes.