## anonymous 2 years ago I need help with finding the inverse of this function, 2x^5+x^3+1.

1. anonymous

$2x^5+x^3+1$

2. anonymous

hmm, technically, the inverse is just swapping about the variables, now the simplification, well, not so simplistic

3. anonymous

exactly, I know the concept it's just that I have to show my work to simplify and insert a value into the new function.

4. klimenkov

Interesting... Can somebody find the inverse function for $$y=x^2+2x-3$$ ?

5. anonymous

yes... but you will have to restrict the domain.

6. klimenkov

Two branches will satisfy: $-1\pm\sqrt{4+x}.$Now how to solve this one? $2y^5+y^3+1-x=0.$

7. anonymous

throwing it into wolframalpha gives this result http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427ef49hilhcgk

8. klimenkov

This means that the inverse function may not be found.

9. anonymous

whaaaaaat...... But It's in my textbook and part of my homework! Ugh...

10. anonymous

what are you studying?

11. anonymous

calc one... this is the second day of homework.

12. anonymous

you sure it's a 5th power?

13. anonymous

sorry, that's the class. I'm studying Electrical Engineering. just a second and I'll upload a snapshot of the problem

14. anonymous

#74

15. klimenkov

Hahaha. Lol.

16. anonymous

hmm well well

17. anonymous

dude... it says f^-1(a) and gives you the value of a. inverse functions work like this... plug in the y from a function into its inverse and you get out x. compute f(a) for your functions and then f^-1(f(a)) will just be a.

18. anonymous

I know, and I have the -71 as the answer, but it also asks to write down the function which I can't find to save my life!

19. anonymous

oops... i misread. but i doesn't ask for the inverse function... only f^-1(a)

20. anonymous

OH! oh oh oh oh.... Damn....

21. anonymous

if you find a = f(x) then the x that satisfies this will be f^-1(a)

22. DebbieG

Wow that's a lot easier than finding $$f^{-1}(x)$$ :)

23. anonymous

yeah, it's -71...

24. anonymous

craaaaap.... well there went an hour I'll never get back. thanks for all your help guys

25. anonymous

no worries... maybe post the question from your book and your interpretation so you can get to the heart of the sunrise quicker.

26. klimenkov

So, what is an answer?

27. anonymous

it's not -71

28. anonymous

no?

29. anonymous

x = -1 -2 = 2x^5 + x^3 +1 => 2x^5 + x^3 +3 = 0 the only real solution is x = -1

30. anonymous

yeah I get what you mean now... but isn't that just f(-2) now? not the inverse?

31. anonymous

no... f(-1) = -2 so f^-1(-2) = -1

32. DebbieG

Remember that $$f^{-1}(-2)$$ is that domain element x that gives you $$f(x)=-2$$ The function maps the domain to the range. The inverse function maps the range back to the domain.

33. klimenkov

Let $$f^{-1}(a)=x$$. Now map $$f$$ to both of the parts of the equation and use $$f(f^{-1}(a))=a$$. It will become$a=f(x)$Solve this and you will obtain an answer.

34. DebbieG

So by setting $$f(x)=y=-2$$ and solving, you find the x in the domain that takes you to y in the range.

35. anonymous

f^-1(f(x)) = x

36. anonymous

ohhh okay, so put the variable as the solution and work backwards

37. anonymous

they give you a = f(x) and want you to find f^-1(a) = f^-1(f(x)) = x so setting f(x) = a and solving for x gives you what you need.

38. anonymous

make sense?

39. anonymous

Yep :) thanks again y'all

40. anonymous

you're welcome!

41. anonymous

*lightbulb turns on* I think the rest of this will be much easier now

42. anonymous

awesome! let's turn the dark into light!