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Math Help - Showing the existence of solutions

  1. #1
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    Showing the existence of solutions

    Show that the equation
    x^2*e^mod(x)
    Has at least two solutions for x in
    -1<=x<=1.

    Thanks
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  2. #2
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    Quote Originally Posted by alpha View Post
    Show that the equation
    x^2*e^mod(x)
    Has at least two solutions for x in
    -1<=x<=1.

    Thanks
    you do not have an equation here.
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  3. #3
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    Sorry. I made a typing error. Here is the corrected version.
    Thank you for your help.

    Show that the equation
    x^2*e^mod(x)=2
    Has at least two solutions for x in
    -1<=x<=1.
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  4. #4
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    The expression e^{\text{mod}(x)} is meaningless.
    Should it be e^{|x|}~?
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  5. #5
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    Quote Originally Posted by Plato View Post
    The expression e^{\text{mod}(x)} is meaningless.
    Should it be e^{|x|}~?

    Yes. That is what I mean. Sorry, I did not know how to get that symbol on here.

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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by alpha View Post
    Sorry. I made a typing error. Here is the corrected version.
    Thank you for your help.

    Show that the equation
    x^2*e^mod(x)=2
    Has at least two solutions for x in
    -1<=x<=1.
    Note that this is the same as showing x^2 e^{|x|} - 2 = 0 at least twice in the given range.

    Let f(x) = x^2 e^{|x|} - 2 (note that it is a continuous function) and consider f(-1),~f(0), \text{ and }f(1) and apply the intermediate value theorem.
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  7. #7
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    Quote Originally Posted by alpha View Post
    Sorry. I made a typing error. Here is the corrected version.
    Thank you for your help.

    Show that the equation
    x^2*e^mod(x)=2
    Has at least two solutions for x in
    -1<=x<=1.
    let f(x) = x^2 \cdot e^{|x|} - 2

    f(-1) = e-2 > 0

    f(0) = -2 < 0

    f(1) = e-2 > 0

    now use the Intermediate Value Theorem and the fact that f(x) is continuous over the given interval.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by alpha View Post
    Yes. That is what I mean. Sorry, I did not know how to get that symbol on here.

    Thanks
    it is on your keyboard. hold down shift and press the "\" key. | should show up

    otherwise, type "abs" instead of "mod", "mod" means something else in math... but it's better and easier to use the | key
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  9. #9
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    I have no idea of how to apply this to the intermediate value theorem. Can you show me how to answer this question.

    Thank you.
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by alpha View Post
    I have no idea of how to apply this to the intermediate value theorem. Can you show me how to answer this question.

    Thank you.
    what does the theorem say? try to apply it, try something! this "I have no idea how to..." line is starting to wear thin. try!
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  11. #11
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    Well I ahve found out that the theorem is as follows:

    If (x) is a continuous real-valued function on the closed interval from a to b, then, for any y between the least upper bound and the greatest lower bound of the values of , there is an x between a and b with (x) = y

    Now how do I apply this to the question.
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  12. #12
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    I think i have just worked it out.

    For f(x) = x^2 * e^abs(x) -2 f(-1) = e - 2 and f(0) = -2 and f(1) = e - 2

    Since e - 2 > 0, f(x) changes sign between -1 and 0 and again between 0 and 1.

    So there are two solutions to f(x) = 0.

    Is this correct?
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  13. #13
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    Quote Originally Posted by alpha View Post
    I think i have just worked it out.

    For f(x) = x^2 * e^abs(x) -2 f(-1) = e - 2 and f(0) = -2 and f(1) = e - 2

    Since e - 2 > 0, f(x) changes sign between -1 and 0 and again between 0 and 1.

    So there are at least two solutions to f(x) = 0.

    Is this correct?
    ...
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  14. #14
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by alpha View Post
    I think i have just worked it out.

    For f(x) = x^2 * e^abs(x) -2 f(-1) = e - 2 and f(0) = -2 and f(1) = e - 2

    Since f(1) = f(-1) = e - 2 > 0, and f(0) = -2 < 0, f(x) changes sign between -1 and 0 and again between 0 and 1.

    So there are at least two solutions to f(x) = 0.

    Is this correct?
    yes!

    you have the right idea. you just have to tighten it up a bit. you have to mention that the function is continuous and that you are applying the intermediate value theorem (twice), things like that.
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