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F(x) is differentiable at x =1, find f(x)

  1. Nov 5, 2012 #1
    1. The problem statement, all variables and given/known data

    suppose a function f is differentiable at x = 1 and

    lim[h → 0] [itex]\frac{f(1 + h)}{h}[/itex] = 5

    Finf, f(1) AND f'(1)




    3. The attempt at a solution

    f(x) is differentiable at x = 1

    f(x) is continious at x = 1

    lim[x→1] f(x)= f(1)

    f'(x) = lim[h→0] [itex]\frac{f(1 + h) - f(1)}{h}[/itex]

    = 5.lim[h→0]f(1)

    = 5.f(1)

    = 5lim[x→1]f(x)

    Is this correct? I can't see anything else to do.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 5, 2012 #2
    No, that's completely wrong :)

    [tex]f'(1) = \lim_{h\rightarrow 0} \frac{f(1+h)-f(1)}{h} = \lim_{h\rightarrow 0} \frac{f(1+h)}{h}-\lim_{h\rightarrow 0} \frac{f(1)}{h} [/tex]

    You know what the first term is; it's 5. Now you need to constrain f(1) by the fact that f is differentiable at 1.
     
  4. Nov 5, 2012 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The denominator goes to 0 so in order that the limit of the fraction exist at all, the numerator must be what?

    And [itex]f'(1)= \lim_{h\to 0} (f(1+ h)- f(1))/h[/itex]. And now that you know what f(1) is, that limit is precisely what you were given!

    Yes, you are correct to here.

    I don't see how this follows from the above. You can break the above into
    [itex]\lim_{h\to 0} f(1+h)/h- \lim_{h\to 0} f(1)/h= 5- \lim_{h\to 0} f(1)/h[/itex] provided that last limit exists.

    No, that's wrong because of what I wrote before.

     
  5. Nov 5, 2012 #4

    lurflurf

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    Homework Helper

    Consider
    f(x)=a+b(x-1)
    then generalize
     
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