## emma97 What is the inverse of f if…. 2 months ago 2 months ago

1. emma97

$f(3) \sqrt[3]{x-2}$

2. bibby

$f(x)=\sqrt[3]{x-2}$ ?

3. emma97

$f(3)=\sqrt[3]{x-2}$

4. emma97

i forgot to put an = sign the first time

5. bibby

are we finding f'(3) either way to get the inverse, let's replace f(x) with y $y = \sqrt[3]{x-2}$ switch x and y $x = \sqrt[3]{y-2}$ cube both sides $x^3 = {y-2}$ $x^3+2=y$

6. emma97

$x^{3}+2=y^{3}$ is that right?.. i have no idea I'm so stressed

7. bibby

How'd you get y^3? Remember that square roots and squares are inverse operations, right? $(\sqrt{x})^2 = x$ $(\sqrt[3]{x})^3 = x$

8. emma97

oh i got confused i thought you said for ME to cube both sides lol. I'm sorry I'm extremely tired and have to get a bunch of work done by like tomorrow. ughhh. my bad

9. bibby

lol no worries, just always go over your work a bazillion times so that you can prove to yourself that your answer is right. ANYHOW, the inverse is now f(x) =x^2+2 f(3) = 3^2+2 = ??

10. emma97

wait so the answer is f(x) =x^2+2 ?

11. bibby

whoops./ you generally say inverse with a little dash f'(x) so the equation for the inverse is f'(x) = x^2 + 2 then you plug in 3 into the inverse equation (assuming that's the question)

12. emma97

heres the actual question with the answer choices so you can see

13. bibby

god damn. I kept saying x^2 +2 but I meant x^3 + 2

14. emma97

hahah so is it the 3rd option then?

15. bibby

it is, it is...

16. emma97

thank youuu! i have another question! do you mind helping me??

17. bibby

are you gonna post it here or make a new question? I don't mind

18. emma97

ill post it here! and thanks! (:

19. emma97

20. bibby

uh you'd multiply by the conjugate. pardon my handwriting

21. emma97

i dont understand...

22. bibby

to simplify radicals you multiply by the conjugate pair which means flipping the sign

23. bibby

why do you have that answer selected?

24. emma97

what do you mean that answer selected?

25. bibby

26. emma97

the ones with the bubble next to them on the link i posted!

27. bibby

rofl I didn't see it scrolled down a bit.

28. emma97

hahahah its all good! (:

29. emma97

30. myininaya

Perform the operation bibby suggested and then simplify.

31. bibby

you can use wolfram if you don't like learning http://www.wolframalpha.com/input/?i=%28sqrt%283%29-sqrt%286%29%29%2F%28sqrt%283%29%2Bsqrt%286%29%29

32. emma97

its not that i don't like learning its that i am in a hurry and i have no idea whats going on.

33. myininaya

$\frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}+\sqrt{6}} \cdot \frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}-\sqrt{6}}$ Do this multiplication here.

34. myininaya

Why are you in a hurry? Is this a test?

35. emma97

idk how to..

36. emma97

i had to go out of town so I'm behind on a bunch of work due like really soon

37. myininaya

So you don't know how to multiply something like this (a+b)(c+d)?

38. myininaya

A lot of Americans like to use this thing called the foil method.

39. emma97

i have heard of it but idk how to really use it

40. myininaya

41. emma97

all you did was distribute those right?

42. myininaya

|dw:1388451041982:dw|

43. myininaya

yep|dw:1388451059969:dw|

44. myininaya

45. myininaya

So if you had $(\sqrt{3}-\sqrt{6})(\sqrt{3}-\sqrt{6})$ what do you have after multiplying?

46. emma97

idk how to group square roots together or multiply them together

47. myininaya

Do that foil method then we will work on simplifying

48. emma97

hold on its gonna take awhile to plug in

49. myininaya

You don't have to use what I gave you as a formula. Just distribute. Take each of the terms in the first grouping and multiply to each term in the second grouping

50. emma97

oh wow my computer plugged in the wrong numbers so thats not even the right question, this is the right question...

51. myininaya

Well I will show you how to do the first one and you try this one.

52. emma97

i tried solving it and i think its the second answer choice

53. myininaya

$\frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}+\sqrt{6}} \cdot \frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}-\sqrt{6}}$ This was the operation you were asked to perform. I was having you look at the numerator first. So looking at the numerator if we multiply the top out we would get: $\frac{\sqrt{3} \sqrt{3}-\sqrt{3} \sqrt{6}-\sqrt{6} \sqrt{3}+\sqrt{6} \sqrt{6}}{(\sqrt{3}+\sqrt{6})(\sqrt{3}-\sqrt{6})}$

54. myininaya

Do you see how I got that top part?

55. emma97

yes

56. myininaya

If a, b>=0 then $\sqrt{a} \cdot \sqrt{b} =\sqrt{a \cdot b}$

57. myininaya

so basically $\sqrt{3} \cdot \sqrt{3} =3 \text{ and } \sqrt{3} \cdot \sqrt{6}=\sqrt{18}=\sqrt{9 \cdot 2} =\sqrt{9} \cdot \sqrt{2}=3\sqrt{2}$

58. myininaya

combine like terms on top and multiply the bottom out i have to go for now

59. emma97

ok.

60. Chooch146

This has to be the longest problem eve

61. Chooch146

ever* lol my name is eve:P

62. emma97

hahaha hey eve! and yah i know…i have a lot of stuff i need help withhh ughh

63. Chooch146

Hai!

64. emma97

hahah whats up!

65. Chooch146

not mch jst listening to music i was about to leave lol

66. emma97

hahaha ok then! lol

67. Chooch146

bye