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Prove f(x) = x^3 + 3^x is a one-to-one function.

  1. Aug 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove f(x) = x^3 + 3^x is a one-to-one function.

    2. The attempt at a solution

    Sum of one-to-one functions is a one-to-one function (I think/dont know how to prove).
    x^3 is one-to-one, 3^x is one-to-one, thus f(x) is one-to-one. Surely there's a more rigorous proof.
     
  2. jcsd
  3. Aug 15, 2012 #2

    arildno

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    Dearly Missed

    Your attempt is wrong. f(x)=x is one-to-one, so is g(x)=1-x. But the sum f+g is not one-to-one.

    Try to answer the following:
    If a function is NOT one-to-one, what do you know occasionally occurs with the function that cannot happen with a one-to-one function?
     
  4. Aug 15, 2012 #3
    Two x values give the same y value. Are you suggesting I use the horizontal line test?
     
  5. Aug 15, 2012 #4
    Hi quadreg, have you studied derivatives yet ?

    Cheers...
     
  6. Aug 15, 2012 #5
    Yes, I tried solving for critical points, but i get f'(x)=3x^2+(3^x)ln3 and can't solve for x where f'(x) = 0
     
  7. Aug 15, 2012 #6
    You only have to prove that the derivative is either always >=0 or <=0 can you do that ?
    Cheers...
     
  8. Aug 15, 2012 #7
    Ohh, that makes sense. 3x^2 is always >=0 and (3^x)ln3 is always >=0, so f'(x)>0.
    Thankyou.
     
  9. Aug 15, 2012 #8
    Exactly, yo've got it :)
     
  10. Aug 15, 2012 #9

    Mark44

    Staff: Mentor

    Mod note: Moving this thread to Calculus & Beyond
     
  11. Aug 15, 2012 #10

    LCKurtz

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    Almost. You have to be sure that if the derivative is allowed to be 0, it isn't zero on any interval. Of course, that isn't a problem in this example.
     
  12. Aug 16, 2012 #11
    Yes, of course, LCKurtz, :) it's a good idea to point that out, thanks :)
     
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