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Math Help - Inverse Functions

  1. #1
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    Inverse Functions

    Find the inverses of the following functions

    o)  y = 3(x - 1)^2, x \geq 1


    x = 3(y - 1)^2

    x = 3(y - 1)(y - 1)

    x = 3(y^2 - 2y + 1)

    x = 3y^2 - 6y + 3

    x - 3 = 3y^2 - 6y

    x - 3 = 3y(y - 2)

    \frac {x - 3}{y - 2} = 3y

    \frac{\frac {x - 3}{y - 2}}{3} = y

     (3)\frac {x - 3}{y - 2} = y

    \frac{3x - 9}{y - 2} = y

    3x - 9 = \frac {y}{y - 2}

    3x - 9 = 1 - \frac {y}{2}

    3x - 9 - 1= \frac {y}{2}

    3x - 10 = \frac {y}{2}

    \frac {3x - 10}{2} = y


    I'm way off. . .

    And textbook answer is y = 1 + \sqrt \frac {x}{3}, x \geq 0
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Macleef View Post
    Find the inverses of the following functions

    o)  y = 3(x - 1)^2, x \geq 1


    x = 3(y - 1)^2

    x = 3(y - 1)(y - 1)

    x = 3(y^2 - 2y + 1)

    x = 3y^2 - 6y + 3

    x - 3 = 3y^2 - 6y

    x - 3 = 3y(y - 2)

    \frac {x - 3}{y - 2} = 3y

    \frac{\frac {x - 3}{y - 2}}{3} = y

     (3)\frac {x - 3}{y - 2} = y

    \frac{3x - 9}{y - 2} = y

    3x - 9 = \frac {y}{y - 2}

    3x - 9 = 1 - \frac {y}{2}

    3x - 9 - 1= \frac {y}{2}

    3x - 10 = \frac {y}{2}

    \frac {3x - 10}{2} = y


    I'm way off. . .

    And textbook answer is y = 1 + \sqrt \frac {x}{3}, x \geq 0
    here's the trick. think of x as if it was just a regular number. then when you expand everything, and move everything to one side, you'd realize that the expression you have is quadratic in y. just use the quadratic formula and you will get the answer
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    Quote Originally Posted by Jhevon View Post
    here's the trick. think of x as if it was just a regular number. then when you expand everything, and move everything to one side, you'd realize that the expression you have is quadratic in y. just use the quadratic formula and you will get the answer
    I solve the expression and got. . .

    x = (3y - 3)(y - 1)

    and I don't know what to do now
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Macleef View Post
    I solve the expression and got. . .

    x = (3y - 3)(y - 1)

    and I don't know what to do now
    as i said, expand everything. you want to get it in the form ay^2 + by + c = 0
    and the expression for c will have x in it

    so you just use the quadratic formula: y = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}
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    Quote Originally Posted by Jhevon View Post
    as i said, expand everything. you want to get it in the form ay^2 + by + c = 0
    and the expression for c will have x in it

    so you just use the quadratic formula: y = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}
    "the expression for c will have x in it" . . .I don't get this part. . .

    and when I tried solving this equation with the quadratic formula, I ended up with. . .

    x = \frac {6 \pm \sqrt{0}}{6}
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Macleef View Post
    "the expression for c will have x in it" . . .I don't get this part. . .

    and when I tried solving this equation with the quadratic formula, I ended up with. . .

    x = \frac {6 \pm \sqrt{0}}{6}
    where did the y go?

    ok, let's take this one step at a time. first step, expand everything as much as possible and bring everything to one side. type what you get
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    Quote Originally Posted by Jhevon View Post
    where did the y go?

    ok, let's take this one step at a time. first step, expand everything as much as possible and bring everything to one side. type what you get
    x = 3(y - 1)(y - 1)
    0 = 3(y - 1)(y - 1) - x
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Macleef View Post
    x = 3(y - 1)(y - 1)
    0 = 3(y - 1)(y - 1) - x
    expand means there should be no brackets left
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    Quote Originally Posted by Jhevon View Post
    expand means there should be no brackets left
    Oh. . .

    Is the following what you mean. . .?

    0 = 3y^2 - 6y + 3 - x
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    Quote:
    Originally Posted by Jhevon View Post
    where did the y go?

    ok, let's take this one step at a time. first step, expand everything as much as possible and bring everything to one side. type what you get
    x = 3(y - 1)(y - 1)
    0 = 3(y - 1)(y - 1) - x
    This will work and is a correct first step on a way of solving this. The next step of this method is to expand the brackets. (EDIT: sorry didn't see new posts. You already got this part)

    However, I think the problem is more sensibly done this way:

    x = 3(y-1)^2
    <br />
\frac x3 = (y-1)^2
    <br />
\pm \sqrt {\frac x3} = y-1
     y = 1 \pm \sqrt {\frac x3}
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by badgerigar View Post
    This will work and is a correct first step on a way of solving this. The next step of this method is to expand the brackets. (EDIT: sorry didn't see new posts. You already got this part)

    However, I think the problem is more sensibly done this way:

    x = 3(y-1)^2
    <br />
\frac x3 = (y-1)^2
    <br />
\pm \sqrt {\frac x3} = y-1
     y = 1 \pm \sqrt {\frac x3}
    well, that was easy

    we need a staples easy button around here
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