## x3_drummerchick one year ago I dont know how to find the inverse of this! Someone help me get it started?? Will give medals: If g(x) = x^2 + 2x with x ≥ −1, find g^−1(15)=.

1. amilapsn

find x in terms of g(x)

2. x3_drummerchick

i just dont know how to start it, we never learned hwoo to do inverse functions with restrictions

3. amilapsn

Here there's a restriction because, without the restriction $$g^{-1}(x)$$ won't be a function...

4. x3_drummerchick

I know up to this point |dw:1442069050216:dw|

5. amilapsn

you're in the right path..

6. x3_drummerchick

it wouldnt be a one to one function, I understand that part. i just dont know how to fnd this inverse. do i factor x^2+2x or...?

7. amilapsn

do you know how to solve quadratic equations?

8. x3_drummerchick

yes

9. x3_drummerchick

|dw:1442069191193:dw|

10. amilapsn

not quite... Can you solve $$x^2+2x-y=0$$ for x?

11. x3_drummerchick

im not sure how to do that one

12. x3_drummerchick

13. amilapsn

yep

14. x3_drummerchick

does i matter that there is a y in the equation? and do i go about doing it with a=1, b=2 and c=1?

15. amilapsn

you've to treat y as a constant..

16. x3_drummerchick

so a=1, b=2, c=-1; is tat why the restriction says that x>=1? cause the square root of it wouldve made the radical a negative; and you cant take the square root of negative number (its imaginary)

17. x3_drummerchick

|dw:1442069827280:dw|

18. x3_drummerchick

wait i messed that up: it should be

19. x3_drummerchick

|dw:1442069945844:dw|

20. x3_drummerchick

|dw:1442070088984:dw|

21. x3_drummerchick

|dw:1442070113092:dw|

22. x3_drummerchick

|dw:1442070133261:dw|

23. x3_drummerchick

so now what do i do, and is that correct? ^

24. x3_drummerchick

@Jhannybean help :(

25. x3_drummerchick

@amilapsn

26. amilapsn

c=-y

27. x3_drummerchick

i dont think this is right.. can you show some work? that doesnt make sense

28. phi

use the quadratic formula, with a=1, b=2, c= -y

29. phi

you should get $x= \frac{-2\pm \sqrt{4+4y}}{2}$ which simplifies to $x= -1 \pm \sqrt{1+y}$

30. phi

now use the constraint that x $$\ge$$ -1 that means we can't use the - sqr, otherwise x will be less than -1

31. x3_drummerchick

so that is the inverse of the function?

32. x3_drummerchick

oh okay t hat makes sense

33. x3_drummerchick

and so now I just plug in the 15 in for x?

34. x3_drummerchick

no wait, dont i have to solve for y first or something?

35. phi

so now you have a function (one value for every y) $x= -1 + \sqrt{1+y}$ we now "switch x and y" because we assume x is the "input" $y = -1 + \sqrt{1+x}$

36. x3_drummerchick

ohhhhh okay

37. phi

now you have $g^{-1}(x) = -1 + \sqrt{1+x}$

38. x3_drummerchick

thank you so much, that makes alot of sense

39. x3_drummerchick

my professor never showed us how to do these kinds -_- totally makes sense though, thanks again!

40. phi

the first trick was realizing you use the quadratic formula and treat -y as the "c"

41. phi

then the second idea is to use x>=-1 to know we don't use the -sqrt as the solution i.e. we use -1 + sqrt()

42. x3_drummerchick

yeah thats what threw me off because weve been doing pproblems where we just have to solve for 'y', but we didnt do a different approach to solving it, i didnt know we could do that with by putting a variable in the quad formula. that definitely threw me through a loophole lol

43. x3_drummerchick

and we cant use -1 because that would make a nonreal number- or imaginary number

44. x3_drummerchick

@phi quick question, how did you simplify under the radical??

45. x3_drummerchick

what did you do in these 2 steps?|dw:1442072831198:dw|

46. x3_drummerchick

|dw:1442072887637:dw|

47. phi

4+4y= 4(1+y) $\sqrt{4(1+y) }= \sqrt{4} \sqrt{1+y}$

48. x3_drummerchick

oh wow, okay, thanks!

49. phi

btw $\sqrt{4+4y} \ne \sqrt{4}+ \sqrt{4y}$ for example, if y=1 we would have $\sqrt{8} \ne 2+2 \\ \sqrt{8} \ne 4$ as a calculator will show

50. x3_drummerchick

so those 2 are completely different, is it because its not held by multiplication?

51. phi

you can't split the sum inside a square root. you are allowed to factor into terms that multiply

52. x3_drummerchick

oh okay