Results 1 to 11 of 11

Math Help - Function within a function problem

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    8

    Function within a function problem

    Hello again! I am again lost on another problem. I wasn't sure what to call this, so hopefully it's still somewhat understandable. Thank you very much in advance!

    The functions f and g are differentiable. For all x, f(g(x)) = x and g(f(x)) = x. If f(3) = 8 and f'(3) = 9, what are the values of g(8) and g'(8)?

    (A) g(8) = 1/3 and g'(8) = -1/9
    (B) g(8) = 1/3 and g'(8) = 1/9
    (C) g(8) = 3 and g'(8) = -9
    (D) g(8) = 3 and g'(8) = -1/9
    (E) g(8) = 3 and g'(8) = 1/9
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Apr 2009
    Posts
    166
    f(g(x)) = x and g(f(x)) = x

    This means that these functions are inverses.

    Can you take it from there?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,367
    Thanks
    41
    Quote Originally Posted by mamori View Post
    Hello again! I am again lost on another problem. I wasn't sure what to call this, so hopefully it's still somewhat understandable. Thank you very much in advance!

    The functions f and g are differentiable. For all x, f(g(x)) = x and g(f(x)) = x. If f(3) = 8 and f'(3) = 9, what are the values of g(8) and g'(8)?

    (A) g(8) = 1/3 and g'(8) = -1/9
    (B) g(8) = 1/3 and g'(8) = 1/9
    (C) g(8) = 3 and g'(8) = -9
    (D) g(8) = 3 and g'(8) = -1/9
    (E) g(8) = 3 and g'(8) = 1/9
    First from g(f(x)) = x sub. x=3. Next differentiate g(f(x)) = x wrt x so g'(f(x))\cdot f'(x) = 1, then subs x= 3 again.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Apr 2009
    Posts
    166
    You're thinking way too much. =p
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Calculus26's Avatar
    Joined
    Mar 2009
    From
    Florida
    Posts
    1,271
    1.

    since g(f(x)) = x

    g(f(3)) = 3

    But f(3) = 8

    so g(8) = 3

    2. f(g(x)) = x


    f and g are inverses so you have a theorem which tells you

    g ' (8) = 1/ f'(3)

    In general g ' (c) = 1/ f ' (a) where f(a) = c

    Or From first principles:

    [f(g(x)] ' = 1 for all x

    But [f(g(x)] ' also equals f ' (g(x)) g ' (x)

    at x = 3 f ' (g(8)) g ' (8) = 1

    f ' (3) *g ' (8) =1

    9* g ' (8) =1

    g' (8) =1/9

    f and g are inverses so you have a theorem which tells you

    g ' (8) = 1/ f'(3)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,367
    Thanks
    41
    Quote Originally Posted by derfleurer View Post
    You're thinking way too much. =p
    Quote Originally Posted by Calculus26 View Post
    1.

    since g(f(x)) = x

    g(f(3)) = 3

    But f(3) = 8

    so g(8) = 3

    2. f(g(x)) = x


    f and g are inverses so you have a theorem which tells you

    g ' (8) = 1/ f'(3)

    In general g ' (c) = 1/ f ' (a) where f(a) = c

    Or From first principles:

    [f(g(x)] ' = 1 for all x

    But [f(g(x)] ' also equals f ' (g(x)) g ' (x)

    at x = 3 f ' (g(8)) g ' (8) = 1

    f ' (3) *g ' (8) =1

    9* g ' (8) =1

    g' (8) =1/9

    f and g are inverses so you have a theorem which tells you

    g ' (8) = 1/ f'(3)
    Apparently me and Calc26
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Apr 2009
    Posts
    8
    Hi everyone! Thank you so much for all your help! I did understand that the functions are inverses, but I forgot the theorem for the derivatives. So thanks for your explanation, Calculus26! I do have one question. Where does

    But [f(g(x)] ' also equals f ' (g(x)) g ' (x)

    come from? Another theorem? I never learned this theorem/rule, so it's a bit tough on understanding this.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Apr 2009
    Posts
    166
    That's just chain rule.

    sin(x^2)' = cos(x^2)(2x)

    And if you understood that they were inverses, you didn't really need to remember any calculus rules. Just know that the slope of the inverse is 1/m and that your domain/range swap. So f(3) = 8 yields f^-1(8) = 3.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor Calculus26's Avatar
    Joined
    Mar 2009
    From
    Florida
    Posts
    1,271
    Danny-- I thought we were paid by the number of words
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Apr 2009
    Posts
    8
    Quote Originally Posted by derfleurer View Post
    That's just chain rule.

    sin(x^2)' = cos(x^2)(2x)

    And if you understood that they were inverses, you didn't really need to remember any calculus rules. Just know that the slope of the inverse is 1/m and that your domain/range swap. So f(3) = 8 yields f^-1(8) = 3.
    Thanks. =] I just understood everything on this page. Somehow that inverse slope rule slipped my mind. Thanks a million everyone!
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,367
    Thanks
    41
    Quote Originally Posted by Calculus26 View Post
    Danny-- I thought we were paid by the number of words
    Don't we wish
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 20
    Last Post: November 27th 2012, 05:28 AM
  2. Replies: 0
    Last Post: October 19th 2011, 04:49 AM
  3. Replies: 4
    Last Post: October 27th 2010, 05:41 AM
  4. Replies: 3
    Last Post: September 14th 2010, 02:46 PM
  5. Greatest integer function (=step function) problem
    Posted in the Math Topics Forum
    Replies: 5
    Last Post: June 7th 2009, 01:43 PM

Search Tags


/mathhelpforum @mathhelpforum