Science:Math Exam Resources/Courses/MATH110/December 2011/Question 10 (b)
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Question 10 (b)
(b) Prove that in any interval (-d, d) (where ), there are infinitely many points c such that .
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When is ?
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We are trying to find points c in an interval (-d, d) such that . To begin, we set the derivative (from part (a)) equal to -1 and isolate the terms with the variable x.
The question is asking us to find an infinite number of points that satisfy this equation. We now use the information from the hint. If we can find a value of x such that , and for the same value of x, , we will have found a solution.
We already know that when
So in order for ,
(Note that x = 0 cannot be a solution because it makes the function undefined.)
Thus we have a possible solution of the form
where n is a non-zero integer. For this list of x-values, . However, we also need in order to satisfy our equation. If we plug our potential solution into the function, we get:
This expression is only equal to 1 if n is even. Thus we must modify our list of solutions to the following:
Where m is an even integer (other than zero).
Because there are an infinite number of even integers, we have found an infinite number of solutions c such that
The last part of the question asks us to show that, given an arbitrary interval (-d, d) for , an infinite number of our list of solutions will fall inside that interval. The key to this section of this problem is noticing that our list of solutions is a fraction, and as the denominator of the fraction can be arbitrarily large (some integer times pi), it can be arbitrarily close to zero. Thus, no matter which d is chosen to be the endpoint of our interval, there will be some even integer, say M, such that
And for all even m larger than M, the value of will also fall between -d and d, meaning we have an infinite number of solutions inside the interval (-d, d).