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anonymous

  • one year ago

PLEASE HELP!!! I HAVE NO IDEA WHAT IM DOING Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x. f(x) = quantity x-8/ x+7. and gx) = -7x - 8/x-1.

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  1. mathstudent55
    • one year ago
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    I don't understand what the functions are. Can you use parentheses to show the numerator and denominator?

  2. UsukiDoll
    • one year ago
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    agree I can't figure it out either. Can you draw the functions?

  3. mathstudent55
    • one year ago
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    Is this f(x)? \(f(x) = \dfrac{x - 8}{x + 7} \) Is g(x) \(g(x) = \dfrac{-7x - 8}{x - 1} \) ?

  4. anonymous
    • one year ago
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    Yesss @mathstudent55

  5. mathstudent55
    • one year ago
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    Ok. Now we can start. Before we start, next time do it this way. Then there won't be any confusion: f(x) = (x - 8)/(x+7) g(x) = (-7x - 8)/(x - 1) Also, you can draw the functions or use the Latex editor.

  6. mathstudent55
    • one year ago
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    Ok. Let's do it. Functions f(x) and g(x) are inverses if f(g(x)) = g(f(x)) = x

  7. mathstudent55
    • one year ago
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    \(f(x) = \dfrac{x - 8}{x + 7}\) f(g(x)) means to evaluate f(x) using g(x) \(g(x) = \dfrac{-7x - 8}{x - 1} \), so we replace x of f(x) with \(\dfrac{-7x - 8}{x - 1}\)

  8. mathstudent55
    • one year ago
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    \(\large f(x) = \dfrac{x - 8}{x + 7} \) \(\large f(g(x)) = f\left(\dfrac{-7x - 8}{x - 1}\right)\) \(\large f(g(x))\Large = \dfrac{\frac{-7x - 8}{x - 1} - 8}{\frac{-7x - 8}{x - 1} + 7}\) You see, where there was an x in the f(x) function, we now have what g(x) is equal to.

  9. mathstudent55
    • one year ago
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    Now we need to simplify that fraction.

  10. mathstudent55
    • one year ago
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    \(\large f(g(x))\Large = \dfrac{\left( \frac{-7x - 8}{x - 1} - 8 \right)(x - 1)}{\left( \frac{-7x - 8}{x - 1} + 7 \right) (x - 1)}\) \(\large f(g(x))\Large = \dfrac{-7x - 8 - 8(x - 1)}{-7x - 8 + 7 (x - 1)}\) \(\large f(g(x))\Large = \dfrac{-7x - 8 - 8x + 8}{-7x - 8 + 7x - 7}\) \(\large f(g(x))\Large = \dfrac{-15x}{-15}\) \(\large f(g(x))\Large = x\)

  11. mathstudent55
    • one year ago
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    That shows that f(g(x)) = x. Now we need to show that g(f(x)) = x

  12. mathstudent55
    • one year ago
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    \(\large g(x) = \dfrac{-7x - 8}{x - 1} \) \(\large f(x) = \dfrac{x - 8}{x + 7} \) Now we replace the x on the right side of the g(x) function with what f(x) is equal to, \(\dfrac{x - 8}{x + 7} \) \(\large g(f(x)) \Large = \dfrac{-7 \left( \frac{x - 8}{x + 7} \right) -8}{ \frac{x - 8}{x + 7} - 1}\) \(\large g(f(x))\Large = \dfrac{\left[ -7 \left( \frac{x - 8}{x + 7} \right) -8 \right](x + 7)}{\left[ \frac{x - 8}{x + 7} - 1 \right](x + 7)} \) \(\large g(f(x))\Large = \dfrac{-7 (x - 8) -8 (x + 7)}{x - 8 - (x + 7)} \) \(\large g(f(x))\Large = \dfrac{-7x + 56 -8x - 56)}{x - 8 - x - 7} \) \(\large g(f(x))\Large = \dfrac{-15x}{- 15 } \) \(\large g(f(x))\Large = x\)

  13. mathstudent55
    • one year ago
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    Now we have also shown that g(f(x)) = x. Since we now know that f(g(x)) = g(f(x)) = x, we have confirmed that the functions f(x) and g(x) are inverses of each other.

  14. anonymous
    • one year ago
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    thank you sooooooo much @mathstudent55

  15. mathstudent55
    • one year ago
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    You are very welcome.

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