## marcelie one year ago help please !!!! find the inverse f(x) 3x^2+5 please check my work

1. marcelie

|dw:1443237553803:dw|

2. steve816

that is some really messed up handwriting xD

3. marcelie

the answer in my textbook says it -

4. steve816

I wish I can help you, but it is too hard to read my friend.

5. marcelie

sorry lol can you solve it ?

6. steve816

But to find the inverse, simply switch the x and y and solve for y.

7. marcelie

k one sec let me rewrite it

8. steve816

subtract 5, divide by 3, square root it.

9. marcelie

|dw:1443237871401:dw|

10. marcelie

|dw:1443238092545:dw|

11. marcelie

why does my textbook answer has the negative infront of the equation

12. steve816

Because you messed up a little bit.

13. marcelie

oh

14. marcelie

im confused ..

15. steve816

Don't be. Life has many confusing moments.

16. marcelie

yeah

17. steve816

But never give up and keep trying until you are successful!

18. marcelie

$f (x ) = 3x ^{2}+5$

19. marcelie

$x= 3y ^{2}+5$

20. steve816

3y^2 = x-4

21. steve816

$y = \frac{ \sqrt{x-5} }{ \sqrt{3} }$

22. marcelie

so do i first divide then square root it ? because i always get confused

23. steve816

You have to rationalize the denominator but I'm too lazy you can do that :)

24. marcelie

isnt the answer that ?

25. steve816

no, you can't have a square root in the denominator.

26. marcelie

mmk bc my solution has that answer and i dont understand why has negative

27. marcelie

28. marcelie

so thats what i did

29. DanJS

$\sqrt{\frac{ a }{ b }}=\frac{ \sqrt{a} }{ \sqrt{b} }$

30. DanJS

recall, when you solve y = x^2 x = +root(y) or x= - root(y)

31. marcelie

oh , so then i have to write -/+

32. marcelie

soo every inverse has to have that -/+ when i take the sq root ?

33. DanJS

no, that is just a property of square roots

34. DanJS

but you have to figure which is the right answer

35. marcelie

oh bc most of the problems ive done are sq roots but my solotuin book doesnt include -/+

36. marcelie

how do i figure that out

37. DanJS

also another thing to note, $\sqrt{y^2} = \left| y \right|$ the square root of y^2 will always be positive, and is really the absolute value of y, not just y

38. DanJS

and a 3rd recall, the absolute value function is defined |dw:1443239752302:dw|

39. DanJS

that is where the negative comes in

40. DanJS

Also as an aside, since you are (x,y) --->(y,x) swapping y and x, The inverse, if there is one, is a reflection over the line y = x ; through the origin slope 1

41. DanJS

notice the intervals they specified, they include all real numbers, but why break it up to that...

42. DanJS

Do they give just the negative root as an answer or both + and - roots depending on which interval you are talking about in the domain