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anonymous

  • one year ago

inverse function property to show that f and g are inverse f(x)= (x-5)/(3x+4) and g(x)= (5+4x)/(1-3x) please help!

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  1. anonymous
    • one year ago
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    I love these questions

  2. Australopithecus
    • one year ago
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    To find the inverse of a function change x to y and change y to x and then solve for y for example the function: f(x) = 3x - 1 to find its inverse we switch variables y = 3x - 1 switch variables: x = 3y - 1 solve for y x + 1 = 3y (x+1)/3 = y therefore the inverse is: \[f^{-1}(x) = \frac{(x+1)}{3}\]

  3. anonymous
    • one year ago
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    O shizzzz I think he has the floor

  4. Australopithecus
    • one year ago
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    Note: Not all functions have an inverse, for a function to have an inverse it must be one-to-one. To check for this property there is a thing called the horizontal line test see: http://www.mathwords.com/h/horizontal_line_test.htm Essentially for a function to be one to one, every input (x value put into the function) must have a unique output (y value provided as a result of the inputted x value out by the function) for example: A function containing the points: (3, 1) and (4, 1) would not be one to one because there are two x values that give the same output

  5. Australopithecus
    • one year ago
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    sorry for not letting you answer this one Kagome9

  6. Australopithecus
    • one year ago
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    Didnt realize you were going to answer it :)

  7. UnkleRhaukus
    • one year ago
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    \[(f\circ g)(x)\\ =f(g(x))\\ =f\left(\frac{5+4x}{1-3x}\right)\\ =\frac{\left(\frac{5+4x}{1-3x}\right)-5}{3\left(\frac{5+4x}{1-3x}\right)+4}\\ =\frac{\left(\frac{5+4x}{1-3x}\right)-5}{3\left(\frac{5+4x}{1-3x}\right)+4}\times\frac{1-3x}{1-3x}\\ =\frac{\left(5+4x\right)-5(1-3x)}{3\left({5+4x}\right)+4(1-3x)}\\ =\frac{5+4x-5+15x}{15+12x+4-12x}\\ =\frac{19x}{19}\\ =x\] \[(f\circ g)(x)=x\iff g= f^{-1}\]

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