Science:Math Exam Resources/Courses/MATH110/December 2011/Question 10 (b)

MATH110 December 2011

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Question 10 (b)
Let $f(x)=x+2x^{2}\sin \left({\frac {1}{x}}\right)$ (b) Prove that in any interval (-d, d) (where $d>0$), there are infinitely many points c such that $\displaystyle f'(c)=-1$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
When is $\sin(x)=0$?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We are trying to find points c in an interval (-d, d) such that $f'(c)=-1$. To begin, we set the derivative (from part (a)) equal to -1 and isolate the terms with the variable x. $\displaystyle -1=1+4x\sin(1/x)-2\cos(1/x)$ $\displaystyle -2=4x\sin(1/x)-2\cos(1/x)$ $\displaystyle -1=2x\sin(1/x)-\cos(1/x)$ 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 $\sin(1/x)=0$, and for the same value of x, $\cos(1/x)=1$, we will have found a solution. We already know that $\sin(x)=0$ when $x=0,\pm \pi ,\pm 2\pi ,\pm 3\pi ,\dots$. So in order for $\sin(1/x)=0$, $x=\pm {\frac {1}{\pi }},\pm {\frac {1}{2\pi }},\pm {\frac {1}{3\pi }},\dots$. (Note that x = 0 cannot be a solution because it makes the function undefined.) Thus we have a possible solution of the form $x={\frac {1}{n\pi }}$ where n is a non-zero integer. For this list of x-values, $\sin(1/x)=0$. However, we also need $\cos(1/x)=1$ in order to satisfy our equation. If we plug our potential solution into the $\cos(1/x)$ function, we get: $\cos \left({\frac {1}{1/(n\pi )}}\right)=\cos(n\pi )$ This expression is only equal to 1 if n is even. Thus we must modify our list of solutions to the following: $x={\frac {1}{m\pi }}$ 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 $f'(c)=-1$ The last part of the question asks us to show that, given an arbitrary interval (-d, d) for $d>0$, 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 $-d<-{\frac {1}{M\pi }}<0<{\frac {1}{M\pi }}<d$ And for all even m larger than M, the value of ${\frac {1}{m\pi }}$ will also fall between -d and d, meaning we have an infinite number of solutions inside the interval (-d, d).

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