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Loser66

  • one year ago

How many solutions are there on \(sec (x) = e^{-x^2}\) Please, help

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  1. Loser66
    • one year ago
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    Another confusing; \(x^2+4x-15=0\) have 2 solutions, but if we apply Decartes rule, we have only 1 real solution. What is wrong? |dw:1438396954345:dw|

  2. Loser66
    • one year ago
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    @ganeshie8

  3. Loser66
    • one year ago
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    Thanks for replying. There are some problems: 1) I open your link and see nothing 2) my link shows it has only 1 solution. https://www.desmos.com/calculator/bnyfuxsluv 3) I would like to know how to solve it by hand.

  4. Loser66
    • one year ago
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    Oh, I see your mistake, it is \(sec (x) = e^{-x^\color{red}{2}}\), while you put \(e^{-x/2}\)

  5. ganeshie8
    • one year ago
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    |dw:1438397782196:dw|

  6. Loser66
    • one year ago
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    @ganeshie8 I got it :)

  7. ganeshie8
    • one year ago
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    for the first problem, you may try showing that \(\sec(x)\ge 1\) and \(e^{-x^2}\le 1\) and use first derivative to show that \(e^{-x^2}\) is decreasing on either side of its "only" maximum value

  8. Loser66
    • one year ago
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    @ganeshie8 I don't get it. Why we have to do that?

  9. ganeshie8
    • one year ago
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    Both are well known graphs, use that knowledge to your advantage in ur proof. The reasoning goes like this : I know that \(e^{-x^2}\) is a bell shaped graph which stays above x axis and below y=1. I also know that \(|\sec x|\ge 1\). So I want use these two facts in my proof (ofcourse you need to prove these facts in ur proof too)

  10. Loser66
    • one year ago
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    I understood @ganeshie8 Thank you so much.

  11. ganeshie8
    • one year ago
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    np, small correction : for the first problem, you may try showing that \(\color{red}{|}\sec(x)\color{red}{|}\ge 1\) and \(e^{-x^2}\le 1\) and use first derivative to show that \(e^{-x^2}\) is \(\color{red}{\text{strictly}}\) decreasing on either side of its "only" maximum value

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