## iop360 3 years ago isolate y for the following equation:

1. iop360

$x= -14e ^{2-y} + y^3 + 7y + 6$

2. ByteMe

i don't think there is an elementary method to do that.

3. iop360

hmm the original equation asks to find the inverse function of $-14e ^{2-x} + x^3 + 7x + 6$

4. iop360

which means swapping y for x and solving for y

5. iop360

and then find the equation of the line perpendicular to this inverse function at x = 14

6. ByteMe

correct... but isolating the y after you reversed it is very difficult. i guess all you could do here is give the function implicitly.

7. ByteMe

can you give the original problem? looks like you need to take derivative implicitly here.

8. iop360

differentiate both sides of the equation, find y' maybe

9. iop360

ok

10. iop360

let $f(x) = -14e ^{2-x} + x^3 + 7x + 6$ find an equation of the normal line to the graph of f-1(x) at the point on the graph where x=14

11. iop360

normal line btw, means the line w/ a negative reciprocal

12. iop360

aka perpendicular. And the answer to this is y = -33x+464

13. ByteMe

do you know how the derivatives (slopes) of inverse functions are related?

14. iop360

maybe could you state it?

15. iop360

i get $\frac{ dy }{ dx } = \frac{ 1 }{ 14e ^{2-y} + 3y^2 + 7 }$

16. iop360

my prof gave a hint saying we had to find a value for x that would make the $e ^{2-x}$ part equal to 0. dont know if that helps tho

17. hartnn

no, need to find the inverse function, u need to find y when x=14 lets say y=2 so u get $$f^{-1}(14)=2$$ then (14,2) is on the graph of x = f-1(y). The slope of that function at that point is what you are looking for.

18. iop360

how do we find$f ^{-1}(14)$

19. ByteMe

since f(2)=14, then f-1(14)=2

20. iop360

oh right

21. iop360

wasnt 2 just used as a hypothetical

22. hartnn

yeah, thats a challenge $$14 = -14e ^{2-x} + x^3 + 7x + 6$$ solve for x to get x=2 is not easy

23. hartnn

i found it by trial and error method

24. iop360

oh ok

25. hartnn

the 1st thing i thought is to eliminate e, by putting x=2 and voila!

26. iop360

oh i see

27. hartnn

so now u have a point, the derivative,can u find the slope of line....and then the equation?

28. iop360

yes that makes it way easier

29. iop360

thanks to both of you

30. hartnn

*point and derivative of inverse function....

31. hartnn

welcome ^_^

32. ByteMe

yw... also remember you're not looking for the tangent but the normal line

33. hartnn

i got this because i have seen a similar question here, http://www.physicsforums.com/showthread.php?t=397539

34. iop360

i see