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Invertible math problem

  1. Dec 10, 2008 #1
    For x E [tex]\Re[/tex], let
    f(x) = 1 + [tex]\int[/tex]et2 dt
    (the interval for this function goes from (0,x) i just didn't know how to put it on the integral.)

    i. Prove that the range of f is [tex]\Re[/tex] (i.e. prove that for every y E [tex]\Re[/tex] there is an x E [tex]\Re[/tex] such that f(x)=y)

    ii. Prove that f: [tex]\Re[/tex] [tex]\rightarrow[/tex][tex]\Re[/tex] is invertible

    iii. Denote the inverse of f by g. Argue that g is differentiable and show that g satisfies the equation
    g'(y) = e-(g(y))2
    for all y E [tex]\Re[/tex]. Show that g is differentiable twice.

    iv. Determine g(1), g'(1), g"(1).

    okay, so for i, i have no idea
    ii, how can you prove that a function is invertible
    iii, i have no idea
    iv, i just find the first derivative and the 2nd derivative and find the values of all of that.

    Please help me out,
    Thank You
     
  2. jcsd
  3. Dec 11, 2008 #2

    tiny-tim

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    Hi tomboi03! :smile:

    (have an ε :wink:)

    Do you mean [tex]f(x)\ =\ 1\ +\ \int_0^xe^{t^2} dt[/tex] ?

    Hint: what are f(-∞) and f(∞)? :wink:
     
  4. Dec 11, 2008 #3
    Re: invertible

    nope, it's still et2

    hahahaha :D hehehe
    Thanks for helping me! i really appreciate it! :D hehehe
     
  5. Dec 15, 2008 #4
    Re: invertible

    Can someone help me with this.. the previous person didn't help me much... thank you
     
  6. Dec 15, 2008 #5

    Mark44

    Staff: Mentor

    Re: invertible

    OK, so apparently [tex]f(x) = \int_0^x et^2 dt[/tex]
    Have you gone so far as to evaluate this integral?
     
  7. Dec 15, 2008 #6
    Re: invertible

    actually it's e^(t^2)....
    my professor has changed it.
     
  8. Dec 15, 2008 #7
    Re: invertible

    Yes, that makes more sense. Which of the questions are you still having problems with?
     
  9. Dec 15, 2008 #8

    tiny-tim

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    mmm … that's what i said! :rolleyes:
    ok … for
    i. Prove that the range of f is R (i.e. prove that for every y E there is an x E such that f(x)=y)​
    … find f(-∞) and f(∞) and describe how it goes from one to the other. :smile:
     
  10. Dec 15, 2008 #9
    Re: invertible

    i'm not sure if i understand how to solve the integral...

    and i was wondering... does invertible mean show that it is bijective?

    I'm confused.
     
  11. Dec 15, 2008 #10

    tiny-tim

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    Hi tomboi03! :smile:

    You don't need to "solve" it … just look at it :smile: … what is its value at ∞?
    Invertible means it has a unique inverse, so if it's onto and invertible, then yes, it's bijective. :wink:
     
  12. Dec 15, 2008 #11
    Re: invertible

    is it infinity? i'm not sure how i can just loook at it... :'(
     
  13. Dec 15, 2008 #12

    tiny-tim

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    :biggrin:

    well, if you draw it, the integral is the area under it (to the right of the y-axis) …

    isn't that obviously ∞?

    ('cos it keeps going up! :rolleyes:)

    ok, now what's f(-∞)? :smile:
     
  14. Dec 15, 2008 #13
    Re: invertible

    isn't it negative infiinity?
     
  15. Dec 15, 2008 #14
    Re: invertible

    so how do you prove that this is invertible?
     
  16. Dec 15, 2008 #15

    tiny-tim

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    Yup! :biggrin:
    Well, it goes from -∞ to ∞, so all you have to prove is that it doesn't go through any level twice. :wink:
     
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