## Destinyyyy one year ago Help with two problems..

1. Destinyyyy

2. Destinyyyy

I was thinking graph A

3. DanJS

yes

4. Destinyyyy

Alright thank you.

5. Destinyyyy

6. Destinyyyy

For this one ive gotten to y=- square root x +13 , y</ 13

7. Destinyyyy

The square root is only over x

8. ahvei

the first one is A let me look at the second one...do you still need help on the second one of are you finished with it

9. Destinyyyy

Still need help on the second one.

10. Destinyyyy

? The second one is not multiple choice. I dont know where you just got C from.

11. Destinyyyy

I need help on the f(x)= (x-13)^2 , x</ 13

12. Destinyyyy

I dont want the answer told to me. Id like to be shown how to solve it.

13. ahvei

oh is it the tyoe of question you will have to type an answer out for??

14. Destinyyyy

Yes? If you look at the picture it shows that the final answer must be typed in... @phi can you help me with this one?

15. Destinyyyy

I can show my work up to where im stuck at if that will help.

16. Destinyyyy

Yes but my example lost me and i'm stuck at y= - square root x +13 , y</ 13

17. Destinyyyy

Im at the part where I found what the restriction and inverse equal.. Restriction = y</ 13 Inverse= y= - square root x +13

18. phi

in the original y= (x-13)^2 because we are squaring , y will always be positive. y>=0 in the "inverse equation" we still want y>= 0 but we renamed y to x and vice versa $y = -\sqrt{x} +13, x \ge 0$ notice we don't want x to be negative (no sqr of negative numbers) so it makes sense as a test, if x=13 in y= (x-13)^2 we get 0 when we "put in" 0 in the inverse we get y= 13 another test x= -1 in y=(x-13)^2 give y= (-1-13)^2 = 196 we put 196 into the inverse y= - sqr(196)+13 = -14 +13 = -1 we get back -1 so it works

19. Destinyyyy

Sorry but I dont really understand all that. This is my second time going this type of problem but first with a square root. The picture attached is what I have written down

20. Destinyyyy

doing*

21. phi

Do you know the domain and range of the original y= (x-13)^2 ?

22. Destinyyyy

Domain of f= Range of f&-1 = (- infinity sign,13]

23. phi

and what is the range?

24. Destinyyyy

Following my example it should be [0, infinity sign) but I have to complete part A to be sure.

25. phi

the domain is 13 to -infinity if we put in 13 we get y=0 if we put in 0 we get y= 13^2= 169 if we put in -1, we get (-14)^2 = 216 so it make sense the range is 0 to +infinity

26. phi

**(-14)^2= 196

27. phi

the inverse function switches the domain and range in other words, the domain of f inverse is the range of f

28. Destinyyyy

Um okay.. That's part C.. I still need help completing part A.

29. Destinyyyy

Yes I know that.

30. phi

you mean $f^{-1}(x)= -\sqrt{x}+13, \ \ x\ge 0$

31. phi

the reason we say x>=0 is because we know the domain of f inverse is the range of f(x) , and that is [0, +infinity) i.e. x>=0

32. phi

Here is a graph

33. Destinyyyy

I followed the example.. f^-1 = - square root x +13 ; x>/0

34. Destinyyyy

I'm confused by one part.. The last step it squares both sides and flipped the sign </ but then the final answer is shown as x>/0

35. Destinyyyy

Thanks for the graph but I didn't need help with that part.. Just part A had me stuck.

36. phi

** The last step it squares both sides and flipped the sign </ but then the final answer is shown as x>/0*** They are unrelated. as you know , the square root (for example) of 4 is 2 and -2 both work. when you are solving for the inverse, when you take the square root you have two possible equations $y= \sqrt{x}+13\\ y= -\sqrt{x}+13$ we have to pick the one that works the restriction x>=0 comes from looking at the range of f(x) and using that as the domain. (the domain are the allowed x values)

37. phi

or perhaps I don't understand your question?

38. Destinyyyy

Um I'll type out what I mean ...

39. Destinyyyy

y</ 13 -square root x +13 </ -square root x </ o Multiply by -1 both sides square root x >/ 0 (square root x) ^2 >/ 0^2 x</ 0 The sign changed on the last one but the answer the sign is > not <

40. Destinyyyy

Why is the sign for the final answer different from the one that it ends with x</0

41. phi

This is how you would do it $y\le 13\\ - \sqrt{x}+13\le 13\\ - \sqrt{x} \le 0 \\ \sqrt{x} \ge 0 \\ x \ge 0$ (notice when you multiply by -1 , you have to swap the <= to >=)

42. phi

If it does not show that, it is a typo

43. Destinyyyy

Yes I noticed that. But my example flipped it back on the last one. Was it an error? It should have stayed as x>/0

44. phi

Definitely x>=0 As a check, notice if you let x be negative in the inverse function, for example x=-1, $y= -\sqrt{-1} +13$ and we can't find the square root of -1, so y is not a real number. That is a good hint we want x>=0

45. phi

if you try to find the square root of -1 on your calculator is will complain

46. Destinyyyy

Okay so it was an error by Pearson.

47. Destinyyyy

Alright thanks

48. phi

a typo. People who look for errors in math books are better at spelling than noticing if an equation makes sense.

49. Destinyyyy

Its an online system. It has a study area with example and practice problems.. Guess ill have to send them an email to fix the typo. Thank you for your help.