anonymous one year ago For the function f(x) = (8-2x)2 ,find f-1 . Determine whether f-1 is a function.

1. anonymous

@Hero

2. anonymous

@Astrophysics

3. anonymous

@UsukiDoll

4. campbell_st

just start by swapping x and y so the original equation is $y = (8 - 2x)^2$ do the swap $x = (8 - 2y)^2$ now may y the subject of the equation

5. anonymous

So should i solve for x? @campbell_st

6. campbell_st

no you should solve for y... 1st step is to take the square root of both sides of the equation...

7. anonymous

y=4− x√ 2,4+ x√ 2

8. campbell_st

well the problem is that the original equation is a function, the inverse will be a relationship... and your solution seems ok...

9. campbell_st

the way you test the inverse is a function is to use the vertical line test after graphing the curve

10. anonymous

@UsukiDoll

11. anonymous

It doesnt match any of my answers

12. campbell_st

so do you have $f^{-1}(x) = \frac{8 - \sqrt{x}}{2}$

13. anonymous

idk i cant see my answer A lol it's probably that one than

14. anonymous

All my other answers are a + not a - and I cant see answer A

15. UsukiDoll

take the one with the + ? Idk.. but @campbell_st is right. you switch the signs of x and y and solve for y

16. anonymous

Is the final answer correct? If so I'll just chose A because everything else is a + after the 8

17. UsukiDoll

?!!!! ok looks like I have to square root both sides or I won't go further lol

18. anonymous

haha I just want to make sure the answer is correct

19. UsukiDoll

$x = (8 - 2y)^2$ [square both sides so we have y by itself] $\sqrt{x}=8-2y$ ok so I'm just gonna stop at this point. I don't want to get busted :P but anyway we need y by itself so I ____________ 8 from both sides

20. anonymous

subtracr

21. UsukiDoll

yes $\sqrt{x}-8=-2y$ so now what do we need to do to get y by itself?

22. anonymous

23. UsukiDoll

not quite. We can't use addition... try again

24. anonymous

divide?

25. UsukiDoll

yes we divide -2 on both sides.

26. UsukiDoll

$\frac{\sqrt{x}-8}{-2}=y$ we can simplify further

27. UsukiDoll

for starters we can split this equation up.... $\frac{\sqrt{x}-8}{-2}=y \rightarrow \frac{-\sqrt{x}}{2}+\frac{-8}{-2}$

28. anonymous

So his answer was right than

29. UsukiDoll

so what's -8/-2 ?

30. anonymous

-4

31. UsukiDoll

a negative divided by a negative turns positive

32. anonymous

4

33. UsukiDoll

34. UsukiDoll

$\frac{-\sqrt{x}}{2}+4$

35. anonymous

ok this is my final question tonight lol

36. anonymous

For the function f(x) = x2 - 12, find (f o f-1)(4)

37. anonymous

do we put 4 in all the x's?

38. UsukiDoll

hmmmm looks like we need to find f^{-1} first and plug that guy into f(x)

39. UsukiDoll

$f(x) =x^2-12$ f(x) is also y $y=x^2-12$ now how do we find the inverse ?

40. anonymous

solve for y?

41. UsukiDoll

that's the second step what is the first step? what needs to be switched?

42. anonymous

the x and y

43. UsukiDoll

yes now we solve for y $x=y^2-12$

44. UsukiDoll

so how do we get y by itself? we need to __________ 12 to both sides

45. anonymous

46. UsukiDoll

$x+12=y^2$ good now we need to ___________ both sides we need a y by itself... it can't have an exponent

47. anonymous

divide?

48. UsukiDoll

no. umm.. what is the square root of y^2 ?

49. anonymous

nothing?

50. UsukiDoll

$\sqrt{y^2} = ?$

51. UsukiDoll

it can't be nothing.. otherwise we won't have y ._.

52. UsukiDoll

maybe exponential form will be easier $\large y^\frac{2}{2}$ what's 2/2 ?

53. anonymous

1

54. anonymous

it would just turn into y

55. UsukiDoll

yes but keep in mind since we took the square root of y^2 we have to take the square root of the left hand side as well $\sqrt{x+12}=y$

56. UsukiDoll

eep $f^{-1}(x) = \sqrt{x+12}$

57. UsukiDoll

now we will place that inverse function into our $f(x) = x^2-12$ but something unusual is going to happen

58. UsukiDoll

oh wait it won't sorry we will still have something ok good whew

59. anonymous

haha ok

60. UsukiDoll

so f o f^{-1} means place f^{-1} inside the f(x) function.. yeah I was just looking at the number and not the variable xD

61. anonymous

62. UsukiDoll

no try placing f inverse in f(x)

63. anonymous

4?

64. UsukiDoll

yeah

65. anonymous

sweetness

66. anonymous

im gonna need for you to be online whenever i am online lol

67. UsukiDoll

$f^{-1}(x) = \sqrt{x+12} , f(x) = x^2-12$ $f(x) \cdot f^{-1}(x)$ $f^{-1}(x) \cdot f(x) = (\sqrt{x+12})^2-12$ $f^{-1}(x) \cdot f(x) = x+12-12$ $f^{-1}(x) \cdot f(x) =x$ when x = 4 $f^{-1}(x) \cdot f(x) =4$

68. anonymous

Thank you for all of your help, I got an A on my hw

69. anonymous

@UsukiDoll

70. UsukiDoll

oh that's awesome! :D