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The derivative of function under constraint

  1. May 12, 2015 #1
    1. The problem statement, all variables and given/known data

    If f(x+y) = f(x)f(y) for all x and y and f(5) = 2, f'(0)=3, then f'(5) is
    a) 5
    b) 6
    C) 0
    d) None of these
    2. Relevant equations

    Don't know which equation to apply. I was thinking of Rolle's Theorem and mean value theorem here but it is not helping.
    3. The attempt at a solution
    Found f(0) = 1
     
  2. jcsd
  3. May 12, 2015 #2

    pasmith

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    You're not actually told that the function is differentiable anywhere other than at zero, so the first thing to do is prove that [itex]f'(5)[/itex] actually exists. Therefore you should start from [tex]
    f'(5) = \lim_{h \to 0} \frac{f(5 + h) - f(5)}{h}.[/tex]
     
  4. May 12, 2015 #3

    HallsofIvy

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    Notice also that, because f(x+ y)= f(x)f(y), f(x+ 0)= f(x)= f(x)f(0) so we must have f(0)= 1. That helps!
     
  5. May 12, 2015 #4
    Is the answer 6?
     
  6. May 12, 2015 #5

    SammyS

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    I think we're waiting for OP to respond.
     
  7. May 13, 2015 #6
    But I have already shown that in my attempt of post 1. How it is helping me?
    Now what should I do?
    I am getting a 0/0 form on right side and L Hopital rule also is not beneficial here.
     
  8. May 13, 2015 #7

    Dick

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    Derivative limits always have a 0/0 form. Use f(x+y)=f(x)f(y).
     
  9. May 13, 2015 #8
    So, I am getting,
    [tex]
    f'(5) = \lim_{h \to 0} \frac{f(5)f( h) - f(5)}{h}.[/tex]
    Now f(5) = 2 and f(0)=1, so f(h) = 1
    Again getting 0/0 form. What I am supposed to do here ?
     
  10. May 13, 2015 #9

    Dick

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    You are given f'(0)=3. What's that as a difference quotient? Halls gave you a hint in post 3.
     
  11. May 13, 2015 #10
    [tex]
    f'(0) = \lim_{h \to 0} \frac{f(0)f( h) - f(0)}{h}.[/tex]

    Now what? Again 0/0 form.
     
  12. May 13, 2015 #11

    Dick

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    Think about it some more. The limit is 3. You are given that. As for what f(0) is you were also hinted that.
     
  13. May 13, 2015 #12
    Okay , applying L Hopital rule getting f'(h) = 3
    Then using in LH rule once again in f'(5) getting 2f'(h) = 6= f'(5)
    Was it supposed to be done by LH rule?
     
  14. May 13, 2015 #13

    Dick

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    No, you don't need LH. And what is f'(h)=3 supposed to mean? f'(0)=3. You haven't thought about this enough.
     
  15. May 13, 2015 #14
    Not,getting
    [tex]
    f'(0) =3= \lim_{h \to 0} \frac{f( h) - 1}{h}.[/tex]
     
  16. May 13, 2015 #15

    Dick

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    Good so far. Now figure out f'(5). It has that expression in it. Don't you see?
     
  17. May 13, 2015 #16
    Yeah f'(5) = 2*3 = 6 by that expression. Thanks Dick.
    Thanks to other people also.
     
  18. May 13, 2015 #17

    pasmith

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    There is a subtle error in the question.

    You can prove that [itex]f(x) = f(1)^x = e^{x \ln (f(1))}[/itex] and differentiate that by the chain rule. The function is completely determined by f(1), but you're given two pieces of information with which to try to find it. They are not consistent.
     
  19. May 13, 2015 #18
    How f(x) = f(1)x ?
    A counter example-
    Putting x = 0
    We get f(0) = f(1)
    Putting x= 5
    We get f(5) = f(1)
    But f(5) ≠ f(0)
    As we are given f(5) = 2 and we know f(0) = 1
     
  20. May 13, 2015 #19

    SammyS

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    No, f(0) = (f(1))0 = 1
     
  21. May 13, 2015 #20
    Oh, so it is
    (f(1))x not f(1)x
    How can we prove f(x) = (f(1))x ?
     
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