## TimmyG Group Title I need help with finding the inverse of this function, 2x^5+x^3+1. one year ago one year ago

1. TimmyG Group Title

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

2. jdoe0001 Group Title

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

3. TimmyG Group Title

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 Group Title

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

5. pgpilot326 Group Title

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

6. klimenkov Group Title

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

7. TimmyG Group Title

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

8. klimenkov Group Title

This means that the inverse function may not be found.

9. TimmyG Group Title

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

10. pgpilot326 Group Title

what are you studying?

11. TimmyG Group Title

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

12. pgpilot326 Group Title

you sure it's a 5th power?

13. TimmyG Group Title

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

14. TimmyG Group Title

#74

15. klimenkov Group Title

Hahaha. Lol.

16. jdoe0001 Group Title

hmm well well

17. pgpilot326 Group Title

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. TimmyG Group Title

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. pgpilot326 Group Title

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

20. TimmyG Group Title

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

21. pgpilot326 Group Title

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

22. DebbieG Group Title

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

23. TimmyG Group Title

yeah, it's -71...

24. TimmyG Group Title

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

25. pgpilot326 Group Title

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 Group Title

27. pgpilot326 Group Title

it's not -71

28. TimmyG Group Title

no?

29. pgpilot326 Group Title

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

30. TimmyG Group Title

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

31. pgpilot326 Group Title

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

32. DebbieG Group Title

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 Group Title

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 Group Title

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. pgpilot326 Group Title

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

36. TimmyG Group Title

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

37. pgpilot326 Group Title

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. pgpilot326 Group Title

make sense?

39. TimmyG Group Title

Yep :) thanks again y'all

40. pgpilot326 Group Title

you're welcome!

41. TimmyG Group Title

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

42. pgpilot326 Group Title

awesome! let's turn the dark into light!