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anonymous

  • one year ago

Will METAL!! Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x. f(x) = the quantity x minus seven divided by the quantity x plus three. and g(x) = quantity negative three x minus seven divided by quantity x minus one.

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  1. phi
    • one year ago
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    ooh. Time to use the equation editor.

  2. anonymous
    • one year ago
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    |dw:1443790834332:dw|

  3. anonymous
    • one year ago
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    @phi finished

  4. phi
    • one year ago
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    the first part f( g(x) ) means everywhere you see x in f(x), replace it with g(x)

  5. anonymous
    • one year ago
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    how exactly do I replace it.

  6. imqwerty
    • one year ago
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    Metal (:

  7. phi
    • one year ago
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    the first step is \[ f( g(x) ) = \frac{g(x) -7}{g(x)+3} \]

  8. phi
    • one year ago
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    but g(x) is also the messy \( \frac{-3x-7}{x-1} \) so we put that expression in for g(x)

  9. phi
    • one year ago
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    btw, for future reference metal (iron or copper) is different from medal (shiny doodad)

  10. phi
    • one year ago
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    everywhere you see g(x) in \[ \frac{g(x) -7}{g(x)+3} \] erase the g(x) and replace it with \[ \frac{-3x-7}{x-1} \] you get a big mess, but it will simplify with some algebra

  11. anonymous
    • one year ago
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    (-3x-7)-7 --------- (x-1)+3

  12. anonymous
    • one year ago
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    ?

  13. phi
    • one year ago
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    \[ \frac{\frac{-3x-7}{x-1}-7}{\frac{-3x-7}{x-1}+3} \]

  14. phi
    • one year ago
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    that is how you replace g(x) with the messy fraction

  15. anonymous
    • one year ago
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    what happens next

  16. phi
    • one year ago
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    to simplify , I would multiply top and bottom by (x-1) \[ \frac{\left(\frac{-3x-7}{x-1}-7\right)}{\left(\frac{-3x-7}{x-1}+3\right)} \cdot \frac{(x-1)}{(x-1)}\]

  17. anonymous
    • one year ago
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    ok so would that cancel out x-1 or...

  18. phi
    • one year ago
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    do the top first: distribute the (x-1) (which means multiply both terms inside by (x-1)

  19. phi
    • one year ago
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    the top is \[ \left(\frac{-3x-7}{x-1}-7\right)(x-1) \\ \frac{-3x-7}{(x-1)}(x-1) -7(x-1) \]

  20. phi
    • one year ago
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    can you simplify the top ?

  21. anonymous
    • one year ago
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    yes, -3x ----- (x-1) x-1 I'm not sure

  22. phi
    • one year ago
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    \[ \frac{-3x-7}{(x-1)}(x-1) \] that is the first term, the (x-1) cancels

  23. anonymous
    • one year ago
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    if the bottom cancels then the top is -3x-7

  24. phi
    • one year ago
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    yes. and the 2nd term is -7(x-1) if you distribute the -7 what do you get ?

  25. anonymous
    • one year ago
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    -3x-7= -7x+7

  26. phi
    • one year ago
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    there is no = sign, just -3x-7+ -7x + 7 (that mess simplified to that) now combine like terms

  27. anonymous
    • one year ago
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    -10x

  28. phi
    • one year ago
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    yes the top (of the original fraction) simplifies to -10x now we do the bottom \[ \left(\frac{-3x-7}{x-1}+3\right)(x-1) \]

  29. anonymous
    • one year ago
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    you said the bottom right?

  30. phi
    • one year ago
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    the (x-1) times the first term will cancel the (x-1) in the botom

  31. phi
    • one year ago
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    in case you lost track, we are doing the bottom of this mess: \[ \frac{\left(\frac{-3x-7}{x-1}-7\right)}{\left(\frac{-3x-7}{x-1}+3\right)} \cdot \frac{(x-1)}{(x-1)} \]

  32. phi
    • one year ago
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    we found that \[ \frac{\left(\frac{-3x-7}{x-1}-7\right)}{\left(\frac{-3x-7}{x-1}+3\right)} \cdot \frac{(x-1)}{(x-1)} = \frac{-10x}{\left(\frac{-3x-7}{x-1}+3\right)(x-1)} \] and we are doing the bottom \[ \left(\frac{-3x-7}{x-1}+3\right)(x-1)\]

  33. phi
    • one year ago
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    and the bottom \[ \left(\frac{-3x-7}{x-1}+3\right)(x-1)\\ \frac{-3x-7}{(x-1)}(x-1)+3(x-1) \]

  34. anonymous
    • one year ago
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    x-1 cancels out so 3-7 is 4 so -3-4.

  35. anonymous
    • one year ago
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    -3x-4

  36. phi
    • one year ago
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    x-1 cancels so you are left with -3x-7 (for the first term)

  37. phi
    • one year ago
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    \[ \frac{-3x-7}{\cancel{(x-1)}}\cancel{(x-1)}+3(x-1) \\ -3x-7 +3(x-1) \]

  38. phi
    • one year ago
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    now distribute the 3 in the 2nd term

  39. anonymous
    • one year ago
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    -3x-7+3x-3

  40. anonymous
    • one year ago
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    -10

  41. phi
    • one year ago
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    ok, so the bottom simplifies to -10 we found that \[ f(x) = \frac{\frac{-3x-7}{x-1}-7}{\frac{-3x-7}{x-1}+3} = \frac{-10x}{-10} \] one more step to go

  42. phi
    • one year ago
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    * f( g(x) ) =

  43. anonymous
    • one year ago
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    1?

  44. phi
    • one year ago
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    you can't ignore the x remember if you have \[ \frac{-10 \cdot x }{-10} \] that is the same as \[ \frac{-10}{-10} \cdot \frac{x}{1} \]

  45. anonymous
    • one year ago
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    1x

  46. phi
    • one year ago
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    yes or just x you just showed f( g(x) ) = x the question was ***Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x. *** you just did the first part. now they want to do it the "other way" show g( f(x) ) = x I hope you are a "high-energy" person, because it's more work.

  47. anonymous
    • one year ago
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    is this right?

  48. phi
    • one year ago
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    it's almost right, g(x) is (-3x-7)/(x-1)

  49. phi
    • one year ago
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    it can get confusing unless we are slow and methodical \[ g(x) = \frac{-3x-7}{x-1} \\ g( f(x)) = \frac{-3f(x)-7}{f(x) -1} \]

  50. phi
    • one year ago
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    now replace f(x) with its expression \[ g( f(x)) = \frac{-3\frac{x-7}{x+3}-7}{\frac{x-7}{x+3} -1} \]

  51. phi
    • one year ago
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    follow?

  52. anonymous
    • one year ago
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    um not really

  53. phi
    • one year ago
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    do you follow this part \[ g(x) = \frac{-3x-7}{x-1} \\ g( f(x)) = \frac{-3f(x)-7}{f(x) -1} \]

  54. anonymous
    • one year ago
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    yes

  55. phi
    • one year ago
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    and f(x) is "short-hand" (or the name for) \[ \frac{x-7}{x+3} \] so everywhere we see f(x), we can put in the "long form"

  56. phi
    • one year ago
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    \[ g( f(x)) = \frac{-3f(x)-7}{f(x) -1} \] the top says "multiply -3 times f(x)" then subtract 7 in math, and replacing f(x) with its expression, we would write -3 * (x-7)/(x+3) then subtract 7: -3 * (x-7)/(x+3) -7 \[ -3 \cdot \frac{x-7}{x+3} - 7\]

  57. anonymous
    • one year ago
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    oh yea cancel out x-3 right

  58. anonymous
    • one year ago
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    x+3

  59. phi
    • one year ago
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    we haven't gotten that far, but that will be the idea. First, do you see how we get \[ g( f(x)) = \frac{-3\frac{x-7}{x+3}-7}{\frac{x-7}{x+3} -1} \]

  60. anonymous
    • one year ago
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    yes I see that now.

  61. phi
    • one year ago
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    now you see the (x+3) that would be nice to get rid of. multiply top and bottom by (x+3)

  62. anonymous
    • one year ago
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    yes

  63. phi
    • one year ago
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    \[ \frac{\left(-3\cdot \frac{x-7}{x+3}-7\right)(x+3)}{\left(\frac{x-7}{x+3} -1\right)(x+3)} \]

  64. phi
    • one year ago
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    can you do the top ? post your steps. the first step is "distribute" the (x+3)

  65. anonymous
    • one year ago
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    I thought x+3 cancels out?

  66. phi
    • one year ago
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    distribute means multiply each term inside the parens for example: (a + b)(x+3) = a(x+3) + b(x+3) use that same rule on \[ \left(-3\cdot \frac{x-7}{x+3}-7\right)(x+3) \]

  67. anonymous
    • one year ago
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    -3(x-7)-7 -3x+21-7 -3x+14 ?

  68. phi
    • one year ago
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    "terms" are things multiplied together. the -3 * (x-7)/(x+3) is all one term terms are separated by + or -. in other words, the -7 is also a term Here is what it looks like, after distributing the (x+3) \[ \left(-3\cdot \frac{x-7}{x+3}-7\right)(x+3)\\ -3\cdot \frac{x-7}{x+3}\cdot (x+3) -7 \cdot (x+3) \] notice the -7 is also multiplied by (x+3)

  69. phi
    • one year ago
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    you did the first part correctly. -3(x-7) but you also have -7(x+3)

  70. anonymous
    • one year ago
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    -3x+14-7x-21 4x-7

  71. anonymous
    • one year ago
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    -10x-7

  72. phi
    • one year ago
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    -3(x-7) + -7(x+3) try again

  73. anonymous
    • one year ago
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    -3x-7+-7x+21 -10x+14

  74. phi
    • one year ago
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    ok but how are you doing -3(x-7) ?

  75. anonymous
    • one year ago
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    ugh I messed up again. -3x+21+-7x-21

  76. anonymous
    • one year ago
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    -10x

  77. phi
    • one year ago
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    yes, that looks good. now we do the bottom \[ \left(\frac{x-7}{x+3} -1\right)(x+3) \]

  78. phi
    • one year ago
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    any luck?

  79. anonymous
    • one year ago
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    no luck :(

  80. phi
    • one year ago
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    can you distribute (x+3) by writing (x+3) next to each term inside the parens?

  81. anonymous
    • one year ago
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    (x-7) -1(x+3)

  82. phi
    • one year ago
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    ok, now distribute the -1

  83. anonymous
    • one year ago
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    x-7-1x-3

  84. anonymous
    • one year ago
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    -10

  85. phi
    • one year ago
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    ok, so we have \[ g( f(x) ) = \frac{-10x}{-10} \]

  86. anonymous
    • one year ago
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    x

  87. phi
    • one year ago
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    yes, so you have shown that f(g(x))= x and g(f(x))= x which proves that f(x) and g(x) are inverses.

  88. anonymous
    • one year ago
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    Thank you soo much :) can we do one more problem it won't take that long.

  89. phi
    • one year ago
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    if you go fast

  90. anonymous
    • one year ago
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    Find the angle θ (if it exists) in the interval [0°, 90°) for which sin θ = cos θ. θ = 30° θ = 45° No such angle exists. θ = 60°

  91. anonymous
    • one year ago
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    options^

  92. phi
    • one year ago
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    divide both sides by cos theta

  93. anonymous
    • one year ago
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    45 degs

  94. phi
    • one year ago
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    yes, you get sin x/cos x = 1 tan x = 1 x= atan 1 = 45º

  95. phi
    • one year ago
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    or you can remember that sin 45 = sqr(2)/2 and cos 45 = the same thing

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