## Loser66 one year ago How many solutions are there on $$sec (x) = e^{-x^2}$$ Please, help

1. 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|

2. Loser66

@ganeshie8

3. 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.

4. Loser66

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

5. ganeshie8

|dw:1438397782196:dw|

6. Loser66

@ganeshie8 I got it :)

7. 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

8. Loser66

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

9. 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)

10. Loser66

I understood @ganeshie8 Thank you so much.

11. 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