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Problem: The equation $x=-5\cos(x)$ has at least $3$ distinction solutions. Use the intermediate value theorem to show that this is true.

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I drew the function,but I don't know what to do next.

user136877
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2 Answers2

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Always with these problems, consider the behavior of $g(x) = x+5 \cos(x)$. Now consider the points $-3, -1, 1,3,5$.

Sandeep Silwal
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I presume you mean it intersects the line $y=x$ at three places. Then, apply IVT as suggested. Guess $f(0) = -5 < 0 $. Now, $f(-\pi/2) = \pi/2 > 0$. Similarly, $f(\pi) = 5 - \pi > 0$ and $f(3\pi/2) = -3\pi/2 < 0$. Now, invoke IVT. We have some root between $-\pi/2$ and $0$; another between $0$ and $\pi$; and yet another between $\pi$ and $3\pi /2$.

Christopher K
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  • what I dont know is how to apply IVT.the text book of mine didn't explain IVT vey well. Do I find the the mid point of the 2 points? – user136877 Apr 16 '14 at 22:55
  • No, what you want to do is take the difference of the two functions (let this be $g$) and show that it is positive at one point, $a$ and then negative at point $b$. By continuity, there exists an intermediary point $c \in (a, b)$ such that $g(c) = 0$. Then, by construction, your two functions intersect at that point. For this type of problem, try to pick values that are easy to calculate (and use the graph, since it is provided to help - that's what I did). – Christopher K Apr 16 '14 at 22:58
  • As you are saying above that there is a root between $-\pi/2$ and 0,how do I find that root? – user136877 Apr 16 '14 at 23:09
  • You don't need to and can't - at least analytically. If $-cos(5x) > x$ at one point but $-\cos(5x) < x$ at another, then there are no jumps or discontinuities how can one curve pass the other, unless they intersect. This is the power IVT; you do not need to explicitly evaluate the roots. – Christopher K Apr 16 '14 at 23:11