Loser66
  • Loser66
How many solutions are there on \(sec (x) = e^{-x^2}\) Please, help
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Loser66
  • 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
@ganeshie8
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Loser66
  • Loser66
Oh, I see your mistake, it is \(sec (x) = e^{-x^\color{red}{2}}\), while you put \(e^{-x/2}\)
ganeshie8
  • ganeshie8
|dw:1438397782196:dw|
Loser66
  • Loser66
@ganeshie8 I got it :)
ganeshie8
  • 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
  • Loser66
@ganeshie8 I don't get it. Why we have to do that?
ganeshie8
  • 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
  • Loser66
I understood @ganeshie8 Thank you so much.
ganeshie8
  • 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.