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Homework Help: Inverse function

  1. Jan 27, 2005 #1
    How do I find the inverse of the following function: f(x)= -3x^2 + 6x + 2
    The answer is +/- sqrt( [2-x/3] +1 ) + 1

    I have no idea how to get that answer. I tried switching x and y and solving for x but I can't get the answer. TiA.
     
  2. jcsd
  3. Jan 27, 2005 #2
    use complete square.....
     
  4. Jan 27, 2005 #3
    I get +/- sqrt(x-5/-3) +1
     
  5. Jan 27, 2005 #4

    dextercioby

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    Can u solve a quadratic equation???

    If so,then solve for [itex]x=x(y) [/itex] from the equation
    [tex]y=-3x^{2}+6x+2 [/tex]

    Or you can complete the square...Correctly...

    Daniel.
     
  6. Jan 27, 2005 #5
    walk me through it... completing the square, I get -3(y-1)^2 + 5 ... and I wrote the inverse of that. I just want to know what I'm doing wrong...
     
  7. Jan 27, 2005 #6
    if your answer read like +/- sqrt((x-5)/-3) +1, then your answer is completely same as the textbook answer... why do you need help?
     
  8. Jan 27, 2005 #7

    dextercioby

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    you mean with "x" as in
    [tex] y=-3(x-1)^{2}+5 [/tex]

    Now solve for "x"...

    Daniel.
     
  9. Jan 27, 2005 #8

    HallsofIvy

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    Strictly speaking, f(x)= -3x2+ 6x+ 2 doesn't have an inverse!

    Set y= -3x2+ 6x+ 2 and then use the quadratic formula to solve the quadratic equation -3x2+ 6x+ (2- y)= 0. You will see a + or - in the formula. In order to get an inverse, we need to restrict the domain. Since the vertex of the parabola is (1, 0.5), if x< 1 we get one inverse (with the -) and if x> 1 we get the other.
     
  10. Jan 27, 2005 #9
    Yeah, I drew the graph... thats why I put +/- ... + is one inverse, - is the other

    ... I see that it's the same now... I dont get why they're different though... I mean I could tack on +1 -1 etc. How was the textbook answer found... (just out of curiosity)?

    TiA
     
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