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- Loser66

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

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- Loser66

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

- katieb

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- Loser66

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|

- Loser66

- Loser66

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.

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- Loser66

Oh, I see your mistake, it is \(sec (x) = e^{-x^\color{red}{2}}\), while you put \(e^{-x/2}\)

- ganeshie8

|dw:1438397782196:dw|

- Loser66

@ganeshie8 I got it :)

- ganeshie8

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

- Loser66

@ganeshie8 I don't get it. Why we have to do that?

- ganeshie8

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)

- Loser66

I understood @ganeshie8
Thank you so much.

- ganeshie8

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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