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Inverse function help

  1. Aug 19, 2016 #1
    1. The problem statement, all variables and given/known data

    Let f(x) = 1−3x−2x^2 , x ∈ [−2, −1]. Use the Horizontal Line Test to show that f is 1–1 (on its given domain), and find the range R of f. Then find an expression for the inverse function f −1 : R → [−2, −1].

    3. The attempt at a solution
    I have already done the horizontal line test but I am unsure about my working out for the other parts below

    would the range just be:

    f(-2)=-1
    f(-1)=2

    y ∈ [−1, 2]

    finding expression for inverse function

    1−3x−2x^2=y
    -2x^2-3x-y+1=0
    using quadratic formula

    x=(3-sqrt(-8y+17)/4 as (3+sqrt(-8y+17)/4 lies outside the range

    Is this correct?
     
  2. jcsd
  3. Aug 20, 2016 #2

    haruspex

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    Sign error, and missing parenthesis. You should have checked whether it was right by substituting the values of x.
    Yes, but how do you justify it?
    Note that f is continuous and 1-1. What does that tell you about turning points?
     
  4. Aug 20, 2016 #3
    x=-(3-sqrt(-8y+17))/4

    when I sub in y=-1 and 2 I get the correct x values

    Is this correct now?

    The question doesn't say anything about justifying the range though
     
  5. Aug 20, 2016 #4

    haruspex

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    Hmm... I don't still. I get 1/2 and -1/2 now.

    Sure, but you were just guessing, yes? How would you justify it to yourself?
     
  6. Aug 20, 2016 #5
    I get a 1/2 and -1/2 when x=-(3+sqrt(-8y+17))/4

    but when x=-(3-sqrt(-8y+17))/4
    i get -1 and 2
     
  7. Aug 20, 2016 #6

    haruspex

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    -(3-sqrt(-8y+17))/4 with y=-1:
    -(3-sqrt(+8+17))/4
    -(3-sqrt(25))/4
    -(3-5)/4
    -(2)/4
     
  8. Aug 20, 2016 #7
    my bad I was entering it in my calculator wrong
    so the answer should be -(3+sqrt(-8y+17))/4
     
  9. Aug 20, 2016 #8

    haruspex

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    Yes. (You needed a calculator for that?)
     
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