## anonymous one year ago find inverse of y=(x+3)^2

1. anonymous

2. Nnesha

two equal signs ?

3. Nnesha

(x+3)^2 or (x-3)^2 ??

4. anonymous

(x+3)^2

5. Nnesha

okay cool same like last one (x+3)^2 is same as (x+3)(x+3) so apply the foil method to expand (x+3)^2 and then replace x with y and y with x

@Nnesha pls help me with a stats question i'm stuck i've already tagged u

7. Nnesha

stat >.>

8. anonymous

so i foiled (x+3)^2 I got x^2+6x+9 so if you switch the x and y you would get x=y^2+6y-9

9. Nnesha

wait a sec i said something wrong

10. dinamix

$f(x)^{-1} =\sqrt{x}-3$for x>3 and $f(x)^{-1} =-\sqrt{x}-3$ for x<3

11. anonymous

how do yo get that answer @dinamix

12. dinamix

@sarahm22

13. dinamix

only i use my mind cuz its easy for me this things

14. dinamix

@Nnesha did u see my answer

15. Nnesha

No.

16. anonymous

what are the steps to getting the answer

17. dinamix

for ops -3 not 3 sorry i type cery fast

18. Nnesha

how he got the answer: take square root both sides

19. Nnesha

that's what he did: replace x with y and y with x 2nd) then take square root both sides to cancel out the square

20. dinamix

$\sqrt{[x]^2} = |x|$

21. dinamix

did u know this rule right ?

22. dinamix

@Nnesha

23. Nnesha

yes ?

24. dinamix

$\sqrt{(x)^2} = |x|$ did u know this rule @Nnesha

25. Nnesha

yes. I said NO .cuz i didn't know how you got +3

26. dinamix

$y = (x+3)^2---> \sqrt{y}=|x+3| --> x=\sqrt{y}-3 --> f(x)^{-1} = \sqrt{x}-3$

27. dinamix

@Nnesha

28. dinamix

this for only when x>-3

29. dinamix

did u see my steps ?

30. Nnesha

yes $\huge\rm \sqrt{ x}=\pm \sqrt{(y+3)^2}$ just in case if he isn't familiar with the absolute value

31. dinamix

so my method is wrong @Nnesha o.O ?

32. Nnesha

i didn't say that :=)

33. dinamix

so what solution @Nnesha i think i am right cuz no mistakes

34. Nnesha

yes i think so. good job and thanks

35. dinamix

u know how do it for x<-3 right ?@Nnesha

36. dinamix

@Nnesha

37. Nnesha

yes i do

38. dinamix

@freckles my method is right ? look my answer plz

39. dinamix

its not 3 i forget - okey

40. freckles

I guess he was looking for inverse relation since an inverse function doesn't exist. And yeah... $y=(x+3)^2 \\ \text{ take square root of both sides } \\ \pm \sqrt{y}=x+3 \\ \text{ now if } x+3 \text{ is positive or zero } (x+3 \ge 0) \\ \text{ then } x+3=\sqrt{y} \\ \text{ and if } x+3 \text{ is negative } \\ \text{ then } x+3=-\sqrt{y}$ $\text{ solving each } \\ x=\sqrt{y}-3 \text{ if } x+3 \ge 0 \text{ or solving the inequality we get } x \ge -3 \\ x=-\sqrt{y}-3 \text{ if } x+3 <0 \text{ or solving that inequality we get } x<-3 \\ \text{ so we have } y=\sqrt{x}-3 \text{ if } x \ge -3 \\ \text{ or } y=-\sqrt{x}-3 \text{ if } x<-3$ which you already said

41. Nnesha

i didn't say tht ur answer is incorrect ....

42. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @dinamix its not 3 i forget - okey $$\color{blue}{\text{End of Quote}}$$ $$\color{blue}{\text{Originally Posted by}}$$ @dinamix for ops -3 not 3 sorry i type cery fast $$\color{blue}{\text{End of Quote}}$$ i know.

43. dinamix

lol why do that @Nnesha