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Inverse functions and tangent lines

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data
    2 problems, i solved both of them but im not 100 % im right

    Find all points on the curve y=x-2cosx where the tangent line to the curve is parallel to the line y=x and write an equation of the tangent line at such point

    Let f^-1 be the inverse of the one -to- one function f(x)=x^3+x.

    Find (f^-1)'(10).



    2. Relevant equations
    slope of tangent line=f'(of the point)


    3. The attempt at a solution


    1st.

    This seemed tricky and I did not figure it out till I had a good 11 hour sleep, what I did:

    y=x , parallel lines have same slopes=>slope of tangent line is 1

    y'=1+2sinx

    1+2sinx=1

    x=0+2pi*n

    we found all points on the curve where tangent is parallel to y=x

    lets find y coordinate
    y=0-2cos(0)=-2

    equation of tangent line
    y+2=1(x-0)
    y=x-2

    2nd.

    I think this has a twist in it but I cant find it :( , heres what I did

    f^-1(x)=1/(f'(f^-1(x))

    we need to find f^-1(10)
    since we got f(x) its not hard to do so
    10=x^3+x
    x=2

    f^-1(x)=1/(f'(2))
    f'=3x^2+1
    f^-1(x)=1/13

    this is the answer?

    Please check what I have written, I apologize for such a mess t.t
     
  2. jcsd
  3. Oct 20, 2011 #2

    SammyS

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    Problem #1:
    You are missing some roots for the equation, [itex]\sin(x) = 0\,.[/itex] Look at the unit circle. For what angle(s) is y = 0 ?

    Because you are missing some of the solutions, you are also missing at least one tangent line. Also, why is it that a single tangent line is sufficient for all the values of x that you did find?
     
  4. Oct 20, 2011 #3

    Mark44

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    Actually sin(x) = 0 for x = n*pi, not just multiples of 2pi.
    No. This is just one of the points at which the tangent line has a slope of 1. Your problem statement says "write an equation of the tangent line at such point". It's not clear to me what this means.
     
  5. Oct 20, 2011 #4

    SammyS

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    Problem #2:

    What you have done is correct.

    For exponents and/or superscripts, use the Advanced message screen. (Click on "Go Advanced" below.)

    Once on the advanced message screen, use the "X2" button above the message box.

    Then you can write (f^-1)' (x)=1/(f '(f^-1(x)) as (f -1)' (x)=1/(f  '(f -1(x))
     
  6. Oct 20, 2011 #5
    Yep, you are right, sinx=0, x=pi n , not 2pi n.

    sinx=0

    x=pi * n

    for x=pi

    y=pi+2

    equation of tangent line

    y-(pi+2)=1*(x-pi)

    Well, it asks to write an equation of the tangent line, so I was thinking it was only 1, guess I was wrong.


    But can't I write infinitive amount of tangent lines?since of n?what do I do then?Do I also write for n =-1 ,+-2, +-3?
     
    Last edited: Oct 20, 2011
  7. Oct 20, 2011 #6

    SammyS

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    Prob. #1:

    cos(±π) = -1

    = cos(±3π) = cos(±5π) = cos(±7π) = ...

    Also, the line y = x - 2, is tangent to y = f(x) at all points where x = 2πn, so you luck out. (You probably should show that this is the case.)

    A different line is tangent to y = f(x) for x = odd multiples of π .
     
  8. Oct 20, 2011 #7
    Thanks a lot for your help!!!
     
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