Results 1 to 7 of 7

Math Help - Find the inverse function, if exists..

  1. #1
    Member
    Joined
    Jan 2009
    Posts
    236

    Find the inverse function, if exists..

    hello
    am stopped at this problem -_-

    let f(x)=2x^2-8x where x>2
    Find a formula for the inverse function of it, if exists.

    My solution:
    f`(x)=4x-..... so f is increas..so its 1-1funct..so inverse is exist.

    its easy to prove that the inverse function is exist..=)

    But the problem is finding the inverse function
    ok let y=2x^2-8x ..
    how to find the x in terms of y !!
    i tried to get the natural logarthmic of both sides but it failed :S

    any help?!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,346
    Thanks
    29
    Quote Originally Posted by TWiX View Post
    hello
    am stopped at this problem -_-

    let f(x)=2x^2-8x where x>2
    Find a formula for the inverse function of it, if exists.

    My solution:
    f`(x)=4x-..... so f is increas..so its 1-1funct..so inverse is exist.

    its easy to prove that the inverse function is exist..=)

    But the problem is finding the inverse function
    ok let y=2x^2-8x ..
    how to find the x in terms of y !!
    i tried to get the natural logarthmic of both sides but it failed :S

    any help?!
    Switch x\; \text{and}\; y, complete the square in y and then solve for y.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2008
    Posts
    106
    complete the square on the right hand side to give
    y=2x^2-8x

    \frac{y}{2}=x^2-4x

    \frac{y}{2}=(x-2)^2-4

    I assume you can take it from there!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293
    Quote Originally Posted by TWiX View Post
    hello
    am stopped at this problem -_-

    let f(x)=2x^2-8x where x>2
    Find a formula for the inverse function of it, if exists.

    My solution:
    f`(x)=4x-..... so f is increas..so its 1-1funct..so inverse is exist.

    its easy to prove that the inverse function is exist..=)

    But the problem is finding the inverse function
    ok let y=2x^2-8x ..
    how to find the x in terms of y !!
    i tried to get the natural logarthmic of both sides but it failed :S

    any help?!
    f(x) IS one to one, but only because the domain has been restricted to all x-values greater than the turning point, not for the reason you stated.

    The domain of f is x > 2, so the range is y > 0, since the function is increasing when x>2...

    Note that for an inverse function, the domain and range swap, which implies that the x and y values swap.

    So f^{-1}(x): x = 2y^2 - 8y

    \frac{1}{2}x = y^2 - 4y

    \frac{1}{2}x + \left(-\frac{4}{2}\right)^2 = y^2 - 4y + \left(-\frac{4}{2}\right)^2

    \frac{1}{2}x + 4 = (y - 2)^2

    Can you solve for y now? Remember that the domain of f^{-1} is x>0 since \textrm{dom}f^{-1} = \textrm{ran}f.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2009
    Posts
    236
    Get it now ..
    Need to solve the equation for x without using the complete of the square ??!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Mar 2007
    Posts
    1,240

    Talking

    Quote Originally Posted by TWiX View Post
    let f(x)=2x^2-8x where x>2

    Find a formula for the inverse function of it, if exists.
    Use the standard process:

    Rename "f" as "y": y = 2x^2 - 8x

    Solve this for "x=". In this case, use the Quadratic Formula:

    . . . . . 2x^2\, -\, 8x\, -\, y\, =\, 0

    . . . . . x\, =\, \frac{-(-8)\, \pm\, \sqrt{(-8)^2\, -\, 4(2)(-y)}}{2(2)}\, =\, \frac{8\, \pm\, \,\sqrt{64\, +\, 8y}}{4}

    ...and so forth. In this case, you were given the restriction of x > 2, so you know you need only the positive square root:

    . . . . . x\, =\, 2\, +\, \frac{\sqrt{16\, +\, 2y}}{2}

    Then continue with the usual steps: Swap "x" and "y", and rename the new "y" as f-inverse.

    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jan 2009
    Posts
    236
    Thank you!
    imo i think that solution is more easy than the first one -_-

    Thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Testing if an inverse exists
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: October 31st 2010, 05:13 AM
  2. Replies: 3
    Last Post: August 31st 2010, 09:19 PM
  3. Replies: 1
    Last Post: May 12th 2010, 05:11 AM
  4. find inverse function
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 13th 2009, 08:08 AM
  5. Find the Inverse Function
    Posted in the Calculus Forum
    Replies: 10
    Last Post: October 9th 2009, 01:11 PM

Search Tags


/mathhelpforum @mathhelpforum