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anonymous

  • one year ago

@shalante can you help me with a problem

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  1. anonymous
    • one year ago
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    i just need help graphing something ..

  2. anonymous
    • one year ago
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    Post it

  3. anonymous
    • one year ago
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    okay so it says Using the two functions listed below, insert numbers in place of letters a,b,c and d so that f(x) and g(x) inverses \[f(x) = \frac{ x+a}{ b }\]

  4. anonymous
    • one year ago
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    and g(x) - cx - d

  5. anonymous
    • one year ago
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    for this part, i chose to do this f(x) = 2x + 5

  6. anonymous
    • one year ago
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    to find g(x) we would have to find the inverse of f(x) right?

  7. anonymous
    • one year ago
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    becaue i did it that way, and got this: y=2x+5 (because i replaced f(x) and y)

  8. anonymous
    • one year ago
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    am i correct so far?

  9. anonymous
    • one year ago
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    f(x) has to be (x+any number)/a different number.

  10. anonymous
    • one year ago
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    i got x-5=2y/ 2

  11. anonymous
    • one year ago
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    and then after simplifying that i got y = x-5 / 2

  12. anonymous
    • one year ago
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    then where it says to evaluate g(x) i wasnt sure but this is how i did it:

  13. anonymous
    • one year ago
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    is it g(x)-cx-d or g(x)=cx-d?

  14. anonymous
    • one year ago
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    g(f(x)) = (2x+5)-5 / 2

  15. anonymous
    • one year ago
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    its g(x)=cx-d

  16. anonymous
    • one year ago
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    wait im sorry i didnt include the parts

  17. anonymous
    • one year ago
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    for part 1 it said show the work that the inverse of f(x) is g(x)

  18. anonymous
    • one year ago
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    2nd*

  19. anonymous
    • one year ago
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    the third part says to show my work evaluating g(x) and thats how i got the last ones

  20. anonymous
    • one year ago
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    i got g(f(x)) = (2x+5)-5 / 2 , because i thought i would have to replace f(x) with the X in the g(x) ... if that makes any sense

  21. anonymous
    • one year ago
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    and at the end its asking me to graph it and im lost o_o

  22. anonymous
    • one year ago
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    what do you have to graph? f(g(x))?

  23. anonymous
    • one year ago
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    yes ..

  24. anonymous
    • one year ago
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    i got this far : \[g(f(x)) = \frac{ (2x+5)-5 }{ 2 }\]

  25. anonymous
    • one year ago
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    but im not sure how to graph it now

  26. anonymous
    • one year ago
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    Can you type out the whole problem correctly in one box to make sure?

  27. anonymous
    • one year ago
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    thats it....

  28. anonymous
    • one year ago
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    tell me part 1,2,3 exactly from your assignment

  29. anonymous
    • one year ago
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    I need to know all the parts to do the steps correctly.

  30. anonymous
    • one year ago
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    Task 1 Part 1. Using the two functions listed below, insert numbers in place of the letters a, b, c, and d so that f(x) and g(x) are inverses. f(x)= x+a / b g(x)=cx−d Part 2. Show your work to prove that the inverse of f(x) is g(x). Part 3. Show your work to evaluate g(f(x)). Part 4. Graph your two functions on a coordinate plane. Include a table of values for each function. Include 5 values for each function. Graph the line y = x on the same graph.

  31. anonymous
    • one year ago
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    And if you go to the top, thats how i did the work...

  32. anonymous
    • one year ago
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    Part 1 Let a=5, b=2,c=2,d=5 so |dw:1440132478788:dw| Part 2 |dw:1440132298809:dw| |dw:1440132437924:dw| Part 3:|dw:1440132646061:dw| |dw:1440132696527:dw|

  33. anonymous
    • one year ago
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    Part 4|dw:1440132750951:dw|

  34. anonymous
    • one year ago
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    You are supposed to graph f(x), g(x) and y=x not g(fx))

  35. anonymous
    • one year ago
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    thaaank youuuuuuuuuuuuuuuuuuuuuuu very muchh !

  36. anonymous
    • one year ago
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    You see that f(x) and g(x) are symmetrical to the line y=x?

  37. anonymous
    • one year ago
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    No problem!

  38. anonymous
    • one year ago
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    yes now i understand it ! thanks alot

  39. anonymous
    • one year ago
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    I forgot to write down the 5 values of each function, but I think you know that. All you do is plug in 5 value into x to find y Glad to help!

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is replying to Can someone tell me what button the professor is hitting...

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