1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    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


    User Avatar
    Homework Helper

    then generalize
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook