## 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

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

2. jdoe0001

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

3. TimmyG

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. pgpilot326

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. TimmyG

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. TimmyG

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

10. pgpilot326

what are you studying?

11. TimmyG

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

12. pgpilot326

you sure it's a 5th power?

13. TimmyG

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

14. TimmyG

#74

15. klimenkov

Hahaha. Lol.

16. jdoe0001

hmm well well

17. pgpilot326

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

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

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

20. TimmyG

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

21. pgpilot326

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. TimmyG

yeah, it's -71...

24. TimmyG

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

25. pgpilot326

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

27. pgpilot326

it's not -71

28. TimmyG

no?

29. pgpilot326

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

30. TimmyG

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

31. pgpilot326

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. pgpilot326

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

36. TimmyG

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

37. pgpilot326

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

make sense?

39. TimmyG

Yep :) thanks again y'all

40. pgpilot326

you're welcome!

41. TimmyG

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

42. pgpilot326

awesome! let's turn the dark into light!