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Math Help - Inverse functions problem

  1. #1
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    Inverse functions problem

    I have been given that f(x) = 3x^2 - 4 and x > 0 ... and I have been asked to find the value of f^-1(x) when x > -4 ...

    Let y = 3x^2 - 4, then 3x^2 - 4 - y = 0
    Solving this equation:

    => \frac{4 \pm \sqrt{16 + 12y}}{6}

    = \frac{4 \pm \sqrt{2} \sqrt{4 + 3y}}{6}

    = \sqrt{4 + 3y} or   \frac {\sqrt{4+3y}}{3}

    Either of them is wrong.. The answer should be \sqrt{\frac{1}{3}(x + 4)}
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  2. #2
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    Quote Originally Posted by struck View Post
    I have been given that f(x) = 3x^2 - 4 and x > 0 ... and I have been asked to find the value of f^-1(x) when x > -4 ...

    Let y = 3x^2 - 4, then 3x^2 - 4 - y = 0
    Solving this equation:

    => \frac{4 \pm \sqrt{16 + 12y}}{6} Mr F says: This is the solution (for x) to {\color{red}3x^2 - 4{\color{blue}x} - y = 0}, but that's not the equation you've got ....

    = \frac{4 \pm \sqrt{2} \sqrt{4 + 3y}}{6}

    = \sqrt{4 + 3y} or  \frac {\sqrt{4+3y}}{3}

    Either of them is wrong.. The answer should be \sqrt{\frac{1}{3}(x + 4)}
    The inverse is found by solving x = 3y^2 - 4 for y, where y = f^{-1} (x). Start by adding 4 to both sides.
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  3. #3
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    Quote Originally Posted by struck View Post
    I have been given that f(x) = 3x^2 - 4 and x > 0 ... and I have been asked to find the value of f^-1(x) when x > -4 ...
    You are given a quadratic whose vertex is at (h, k) = (0, -4). You are told to consider only the right-hand half of the parabola; with this restriction, the function passes the Horizontal Line Test, and is invertible.

    From what you know about graphing quadratics, you know that the graph of f(x), for x > 0, is the right-hand half of a right-side up parabola, with domain x > 0 and range y > -4. The inverse will then be the upper half of a rightward-opening sideways parabola, with domain x > -4 and range y > 0.

    To learn the general process for finding an inverse, try here. Then:

    You have:

    . . . . . f(x)\, =\, 3x^2\, -\, 4

    . . . . . y\, =\, 3x^2\, -\, 4

    To solve for "x=", apply the Quadratic Formula:

    . . . . . 0\, =\, 3x^2\, +\, (-4\, -\, y)

    . . . . . x\, =\, \frac{0\, \pm\, \sqrt{(0)^2\, -\, 4(3)(-4\, -\, y)}}{2(3)}

    . . . . . . . . =\, \frac{\pm \sqrt{12(4\, +\, y)}}{6}

    Since you know you need the range to be y > 0, then you know to take only the "plus" part of the "plus, minus" on the square root.
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