emma97 Group Title What is the inverse of f if…. 8 months ago 8 months ago

1. emma97 Group Title

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

2. bibby Group Title

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

3. emma97 Group Title

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

4. emma97 Group Title

i forgot to put an = sign the first time

5. bibby Group Title

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 Group Title

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

7. bibby Group Title

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 Group Title

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 Group Title

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 Group Title

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

11. bibby Group Title

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 Group Title

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

13. bibby Group Title

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

14. emma97 Group Title

hahah so is it the 3rd option then?

15. bibby Group Title

it is, it is...

16. emma97 Group Title

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

17. bibby Group Title

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

18. emma97 Group Title

ill post it here! and thanks! (:

19. emma97 Group Title

20. bibby Group Title

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

21. emma97 Group Title

i dont understand...

22. bibby Group Title

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

23. bibby Group Title

why do you have that answer selected?

24. emma97 Group Title

what do you mean that answer selected?

25. bibby Group Title

26. emma97 Group Title

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

27. bibby Group Title

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

28. emma97 Group Title

hahahah its all good! (:

29. emma97 Group Title

30. myininaya Group Title

Perform the operation bibby suggested and then simplify.

31. bibby Group Title

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 Group Title

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 Group Title

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

34. myininaya Group Title

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

35. emma97 Group Title

idk how to..

36. emma97 Group Title

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

37. myininaya Group Title

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

38. myininaya Group Title

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

39. emma97 Group Title

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

40. myininaya Group Title

41. emma97 Group Title

all you did was distribute those right?

42. myininaya Group Title

|dw:1388451041982:dw|

43. myininaya Group Title

yep|dw:1388451059969:dw|

44. myininaya Group Title

45. myininaya Group Title

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

46. emma97 Group Title

idk how to group square roots together or multiply them together

47. myininaya Group Title

Do that foil method then we will work on simplifying

48. emma97 Group Title

hold on its gonna take awhile to plug in

49. myininaya Group Title

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 Group Title

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

51. myininaya Group Title

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

52. emma97 Group Title

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

53. myininaya Group Title

$\frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}+\sqrt{6}} \cdot \frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}-\sqrt{6}}$ 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 Group Title

Do you see how I got that top part?

55. emma97 Group Title

yes

56. myininaya Group Title

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

57. myininaya Group Title

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 Group Title

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

59. emma97 Group Title

ok.

60. Chooch146 Group Title

This has to be the longest problem eve

61. Chooch146 Group Title

ever* lol my name is eve:P

62. emma97 Group Title

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

63. Chooch146 Group Title

Hai!

64. emma97 Group Title

hahah whats up!

65. Chooch146 Group Title

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

66. emma97 Group Title

hahaha ok then! lol

67. Chooch146 Group Title

bye