Intermediate value theorem problem












1












$begingroup$


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



enter image description here



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






continuity







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  • $begingroup$
    Try applying the intermediate value theorem to the function $f(x)=-5cos(x)-x$.
    $endgroup$
    – user137301
    Apr 16 '14 at 19:58
















1












$begingroup$


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



enter image description here



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






continuity







share|cite|improve this question











$endgroup$












  • $begingroup$
    Try applying the intermediate value theorem to the function $f(x)=-5cos(x)-x$.
    $endgroup$
    – user137301
    Apr 16 '14 at 19:58














1












1








1


2



$begingroup$


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



enter image description here



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






continuity







share|cite|improve this question











$endgroup$




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



enter image description here



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






continuity




continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 16 '14 at 22:52







user136877

















asked Apr 16 '14 at 19:54









user136877user136877

80116




80116












  • $begingroup$
    Try applying the intermediate value theorem to the function $f(x)=-5cos(x)-x$.
    $endgroup$
    – user137301
    Apr 16 '14 at 19:58


















  • $begingroup$
    Try applying the intermediate value theorem to the function $f(x)=-5cos(x)-x$.
    $endgroup$
    – user137301
    Apr 16 '14 at 19:58
















$begingroup$
Try applying the intermediate value theorem to the function $f(x)=-5cos(x)-x$.
$endgroup$
– user137301
Apr 16 '14 at 19:58




$begingroup$
Try applying the intermediate value theorem to the function $f(x)=-5cos(x)-x$.
$endgroup$
– user137301
Apr 16 '14 at 19:58










2 Answers
2






active

oldest

votes


















0












$begingroup$

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 $f(3pi/2) = -3pi/2 < 0$. Now, invoke IVT. We have some root between $-pi/2$ and $0$; another between $0$ and $pi$; and yet another between $pi$ and $3pi /2$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – user136877
    Apr 16 '14 at 22:55










  • $begingroup$
    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).
    $endgroup$
    – Chris K
    Apr 16 '14 at 22:58










  • $begingroup$
    As you are saying above that there is a root between $-pi/2$ and 0,how do I find that root?
    $endgroup$
    – user136877
    Apr 16 '14 at 23:09










  • $begingroup$
    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.
    $endgroup$
    – Chris K
    Apr 16 '14 at 23:11





















0












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I cant quite get the reason of this. Can you plz add more details?
    $endgroup$
    – user136877
    Apr 16 '14 at 20:05






  • 1




    $begingroup$
    How does one readily calculate $cos(3)$, for example? The whole point of IVT is no calculator required.
    $endgroup$
    – Chris K
    Apr 16 '14 at 23:01












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

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 $f(3pi/2) = -3pi/2 < 0$. Now, invoke IVT. We have some root between $-pi/2$ and $0$; another between $0$ and $pi$; and yet another between $pi$ and $3pi /2$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – user136877
    Apr 16 '14 at 22:55










  • $begingroup$
    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).
    $endgroup$
    – Chris K
    Apr 16 '14 at 22:58










  • $begingroup$
    As you are saying above that there is a root between $-pi/2$ and 0,how do I find that root?
    $endgroup$
    – user136877
    Apr 16 '14 at 23:09










  • $begingroup$
    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.
    $endgroup$
    – Chris K
    Apr 16 '14 at 23:11


















0












$begingroup$

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 $f(3pi/2) = -3pi/2 < 0$. Now, invoke IVT. We have some root between $-pi/2$ and $0$; another between $0$ and $pi$; and yet another between $pi$ and $3pi /2$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – user136877
    Apr 16 '14 at 22:55










  • $begingroup$
    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).
    $endgroup$
    – Chris K
    Apr 16 '14 at 22:58










  • $begingroup$
    As you are saying above that there is a root between $-pi/2$ and 0,how do I find that root?
    $endgroup$
    – user136877
    Apr 16 '14 at 23:09










  • $begingroup$
    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.
    $endgroup$
    – Chris K
    Apr 16 '14 at 23:11
















0












0








0





$begingroup$

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 $f(3pi/2) = -3pi/2 < 0$. Now, invoke IVT. We have some root between $-pi/2$ and $0$; another between $0$ and $pi$; and yet another between $pi$ and $3pi /2$.






share|cite|improve this answer











$endgroup$



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 $f(3pi/2) = -3pi/2 < 0$. Now, invoke IVT. We have some root between $-pi/2$ and $0$; another between $0$ and $pi$; and yet another between $pi$ and $3pi /2$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Apr 16 '14 at 22:59

























answered Apr 16 '14 at 22:53









Chris KChris K

2,6931814




2,6931814












  • $begingroup$
    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?
    $endgroup$
    – user136877
    Apr 16 '14 at 22:55










  • $begingroup$
    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).
    $endgroup$
    – Chris K
    Apr 16 '14 at 22:58










  • $begingroup$
    As you are saying above that there is a root between $-pi/2$ and 0,how do I find that root?
    $endgroup$
    – user136877
    Apr 16 '14 at 23:09










  • $begingroup$
    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.
    $endgroup$
    – Chris K
    Apr 16 '14 at 23:11




















  • $begingroup$
    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?
    $endgroup$
    – user136877
    Apr 16 '14 at 22:55










  • $begingroup$
    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).
    $endgroup$
    – Chris K
    Apr 16 '14 at 22:58










  • $begingroup$
    As you are saying above that there is a root between $-pi/2$ and 0,how do I find that root?
    $endgroup$
    – user136877
    Apr 16 '14 at 23:09










  • $begingroup$
    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.
    $endgroup$
    – Chris K
    Apr 16 '14 at 23:11


















$begingroup$
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?
$endgroup$
– user136877
Apr 16 '14 at 22:55




$begingroup$
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?
$endgroup$
– user136877
Apr 16 '14 at 22:55












$begingroup$
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).
$endgroup$
– Chris K
Apr 16 '14 at 22:58




$begingroup$
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).
$endgroup$
– Chris K
Apr 16 '14 at 22:58












$begingroup$
As you are saying above that there is a root between $-pi/2$ and 0,how do I find that root?
$endgroup$
– user136877
Apr 16 '14 at 23:09




$begingroup$
As you are saying above that there is a root between $-pi/2$ and 0,how do I find that root?
$endgroup$
– user136877
Apr 16 '14 at 23:09












$begingroup$
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.
$endgroup$
– Chris K
Apr 16 '14 at 23:11






$begingroup$
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.
$endgroup$
– Chris K
Apr 16 '14 at 23:11













0












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I cant quite get the reason of this. Can you plz add more details?
    $endgroup$
    – user136877
    Apr 16 '14 at 20:05






  • 1




    $begingroup$
    How does one readily calculate $cos(3)$, for example? The whole point of IVT is no calculator required.
    $endgroup$
    – Chris K
    Apr 16 '14 at 23:01
















0












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I cant quite get the reason of this. Can you plz add more details?
    $endgroup$
    – user136877
    Apr 16 '14 at 20:05






  • 1




    $begingroup$
    How does one readily calculate $cos(3)$, for example? The whole point of IVT is no calculator required.
    $endgroup$
    – Chris K
    Apr 16 '14 at 23:01














0












0








0





$begingroup$

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






share|cite|improve this answer









$endgroup$



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







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Apr 16 '14 at 19:57









Sandeep SilwalSandeep Silwal

5,90811337




5,90811337












  • $begingroup$
    I cant quite get the reason of this. Can you plz add more details?
    $endgroup$
    – user136877
    Apr 16 '14 at 20:05






  • 1




    $begingroup$
    How does one readily calculate $cos(3)$, for example? The whole point of IVT is no calculator required.
    $endgroup$
    – Chris K
    Apr 16 '14 at 23:01


















  • $begingroup$
    I cant quite get the reason of this. Can you plz add more details?
    $endgroup$
    – user136877
    Apr 16 '14 at 20:05






  • 1




    $begingroup$
    How does one readily calculate $cos(3)$, for example? The whole point of IVT is no calculator required.
    $endgroup$
    – Chris K
    Apr 16 '14 at 23:01
















$begingroup$
I cant quite get the reason of this. Can you plz add more details?
$endgroup$
– user136877
Apr 16 '14 at 20:05




$begingroup$
I cant quite get the reason of this. Can you plz add more details?
$endgroup$
– user136877
Apr 16 '14 at 20:05




1




1




$begingroup$
How does one readily calculate $cos(3)$, for example? The whole point of IVT is no calculator required.
$endgroup$
– Chris K
Apr 16 '14 at 23:01




$begingroup$
How does one readily calculate $cos(3)$, for example? The whole point of IVT is no calculator required.
$endgroup$
– Chris K
Apr 16 '14 at 23:01


















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