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Destinyyyy

  • one year ago

Help with two problems..

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  1. Destinyyyy
    • one year ago
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  2. Destinyyyy
    • one year ago
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    I was thinking graph A

  3. DanJS
    • one year ago
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    yes

  4. Destinyyyy
    • one year ago
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    Alright thank you.

  5. Destinyyyy
    • one year ago
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  6. Destinyyyy
    • one year ago
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    For this one ive gotten to y=- square root x +13 , y</ 13

  7. Destinyyyy
    • one year ago
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    The square root is only over x

  8. ahvei
    • one year ago
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    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
    • one year ago
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    Still need help on the second one.

  10. Destinyyyy
    • one year ago
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    ? The second one is not multiple choice. I dont know where you just got C from.

  11. Destinyyyy
    • one year ago
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    I need help on the f(x)= (x-13)^2 , x</ 13

  12. Destinyyyy
    • one year ago
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    I dont want the answer told to me. Id like to be shown how to solve it.

  13. ahvei
    • one year ago
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    oh is it the tyoe of question you will have to type an answer out for??

  14. Destinyyyy
    • one year ago
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    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
    • one year ago
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    I can show my work up to where im stuck at if that will help.

  16. Destinyyyy
    • one year ago
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    Yes but my example lost me and i'm stuck at y= - square root x +13 , y</ 13

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

  18. phi
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    doing*

  21. phi
    • one year ago
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    Do you know the domain and range of the original y= (x-13)^2 ?

  22. Destinyyyy
    • one year ago
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    Domain of f= Range of f&-1 = (- infinity sign,13]

  23. phi
    • one year ago
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    and what is the range?

  24. Destinyyyy
    • one year ago
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    Following my example it should be [0, infinity sign) but I have to complete part A to be sure.

  25. phi
    • one year ago
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    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
    • one year ago
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    **(-14)^2= 196

  27. phi
    • one year ago
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    the inverse function switches the domain and range in other words, the domain of f inverse is the range of f

  28. Destinyyyy
    • one year ago
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    Um okay.. That's part C.. I still need help completing part A.

  29. Destinyyyy
    • one year ago
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    Yes I know that.

  30. phi
    • one year ago
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    you mean \[f^{-1}(x)= -\sqrt{x}+13, \ \ x\ge 0\]

  31. phi
    • one year ago
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    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
    • one year ago
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    Here is a graph

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  33. Destinyyyy
    • one year ago
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    I followed the example.. f^-1 = - square root x +13 ; x>/0

  34. Destinyyyy
    • one year ago
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    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
    • one year ago
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    Thanks for the graph but I didn't need help with that part.. Just part A had me stuck.

  36. phi
    • one year ago
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    ** 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
    • one year ago
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    or perhaps I don't understand your question?

  38. Destinyyyy
    • one year ago
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    Um I'll type out what I mean ...

  39. Destinyyyy
    • one year ago
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    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
    • one year ago
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    Why is the sign for the final answer different from the one that it ends with x</0

  41. phi
    • one year ago
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    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
    • one year ago
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    If it does not show that, it is a typo

  43. Destinyyyy
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    if you try to find the square root of -1 on your calculator is will complain

  46. Destinyyyy
    • one year ago
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    Okay so it was an error by Pearson.

  47. Destinyyyy
    • one year ago
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    Alright thanks

  48. phi
    • one year ago
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    a typo. People who look for errors in math books are better at spelling than noticing if an equation makes sense.

  49. Destinyyyy
    • one year ago
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    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.

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