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horsegirl27

  • one year ago

Please help me out!! I have work for the first part of this question, but then I'm confused. When reading it, it seems like parts 2 and 3 are the same, so I would like some clarification of what to do. For part 4, I'm not quite sure either.... Question posted below.

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  1. horsegirl27
    • one year ago
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    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.

  2. horsegirl27
    • one year ago
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    I'm sorry, those functions are messed up, let me fix that.

  3. horsegirl27
    • one year ago
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    f(x)=x+a/b g(x)=cx-d

  4. horsegirl27
    • one year ago
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    So, for part 1, I have f(x)=x+3/5 and g(x)=5x-3. I'm not sure of what I'm doing in parts 2 and 3.

  5. phi
    • one year ago
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    is f(x) \[ f(x) = \frac{x+a}{b} \]?

  6. horsegirl27
    • one year ago
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    yes.

  7. phi
    • one year ago
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    so we could write \[ f(x) = \frac{1}{b} x + \frac{a}{b} \]

  8. horsegirl27
    • one year ago
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    For which part?

  9. phi
    • one year ago
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    just thinking out loud. For part 2, they want you to show the functions are inverses. start with your: f(x)=(x+3)/5 g(x)= 5x-3 if they are inverses then f( g(x) ) = x

  10. phi
    • one year ago
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    f( g(x) ) is short-hand for : everywhere you see "x" in the expression for f(x), erase the x and write g(x) using f(x)=(x+3)/5 g(x)= 5x-3 f( g(x) ) = ( g(x) +3 ) /5

  11. horsegirl27
    • one year ago
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    ok

  12. phi
    • one year ago
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    now we use g(x) = 5x-3 in f( g(x) ) = ( g(x) +3 ) /5 in other words, on the right hand side, replace g(x) with its "definition" 5x-3 can you do that ?

  13. phi
    • one year ago
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    you should get \[ f( g(x) ) = \frac{( 5x-3 +3)}{5} \]

  14. phi
    • one year ago
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    now simplify that. For part 3, do almost the same thing show g( f(x) ) = x

  15. horsegirl27
    • one year ago
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    ok

  16. horsegirl27
    • one year ago
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    And I am so sorry, I was gone for a moment.

  17. horsegirl27
    • one year ago
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    Ok, so that's what I got for that part.

  18. horsegirl27
    • one year ago
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    So for part 3, to I basically just solve it to show g( f(x) ) = x?

  19. horsegirl27
    • one year ago
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    Because I worked it out and got it. Is that what I want to turn in for part 3, my work?

  20. phi
    • one year ago
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    I assume for part 2, you got f( g(x))= x after simplifying?

  21. phi
    • one year ago
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    \[ f( g(x) ) = \frac{( 5x-3 +3)}{5} = \frac{5x+0}{5} = \frac{5x}{5} = x \]

  22. horsegirl27
    • one year ago
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    yes. That's my work exactly.

  23. phi
    • one year ago
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    for part 3 g( f(x) ) start with f(x)=(x+3)/5 g(x)= 5x-3 g(f(x)) means replace x in 5x-3 with f(x) what do you get ?

  24. horsegirl27
    • one year ago
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    g(f(x))=5(x+3/5)-3 Then the 3's will cancel each other out, and so will the 5's, to get g(f(x))=x.

  25. phi
    • one year ago
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    you have to remember order of operations. \[ g(f(x)) = 5 \cdot \frac{(x+3)}{5} -3 \] multiply before add/subtract

  26. horsegirl27
    • one year ago
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    oh, sorry

  27. horsegirl27
    • one year ago
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    So then it will x+3/1 or 0? And then after that, the 3's will still cancel out, right?

  28. phi
    • one year ago
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    the 5/5 becomes 1 and you get \[ g(f(x)) = 5 \cdot \frac{(x+3)}{5} -3 \\ = 1\cdot (x+3) -3 \\=\ (x+3)-3 \\ =x+3-3 \\=x\]

  29. phi
    • one year ago
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    just like \[ \frac{15}{5} = \frac{5 \cdot 3 }{5} = \frac{5}{5} \cdot 3 = 1 \cdot 3 = 3\]

  30. horsegirl27
    • one year ago
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    ok, that makes sense.

  31. horsegirl27
    • one year ago
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    So for part 2, I'm showing for for f(x), and for part 3, showing work for g(x)? Then graphing? And I'm confused how I should graph this

  32. phi
    • one year ago
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    Include a table of values for each function. Include 5 values for each function. that means they want a table of x y values where y is f(x) or g(x) depending on which function you are doing.

  33. horsegirl27
    • one year ago
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    ok, so for x I could use something like 0, 1, 2, 3, -1 and for y -3, 2, 7, 12, -8? (i might not use those numbers, I just chose them randomly)

  34. phi
    • one year ago
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    Start with the first function f(x) f(x)=(x+3)/5 I would pick small negative and positive numbers for x to get integer answers for f(x) (i.e. y) for example if you start at x= -8, and make x go up by 5 , you will get nice numbers for f(x)

  35. horsegirl27
    • one year ago
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    ok, thanks.

  36. horsegirl27
    • one year ago
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    By the way, any suggestions for an online graph?

  37. phi
    • one year ago
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    I downloaded geogebra.

  38. horsegirl27
    • one year ago
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    Okay, I'll download it too.

  39. horsegirl27
    • one year ago
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    So, graph 5 points for each?

  40. phi
    • one year ago
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    First write down the table x | f(x) -------- -8 | -1 -3 | 0 etc and for g(x) you can pick x= -2 to +2 by 1 x | g(x) -------- -2 | -13 -1 | -8 etc

  41. phi
    • one year ago
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    if you use geogebra you type in f(x)= (x+3)/5 and it will plot the line

  42. horsegirl27
    • one year ago
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    ok

  43. horsegirl27
    • one year ago
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    Got it, thanks!

  44. horsegirl27
    • one year ago
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    I have one more question, want me to make a new thread?

  45. phi
    • one year ago
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    Here is the plot

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  46. horsegirl27
    • one year ago
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    thank you :D

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