1. Not finding help here? Sign up for a free 30min 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!

Question on Mean Value Theorem

Tags:
  1. Aug 14, 2015 #1

    Titan97

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    Let ###f## be double differentiable function such that ##|f''(x)|\le 1## for all ##x\in [0,1]##. If f(0)=f(1), then,
    A)##|f(x)|>1##
    B)##|f(x)|<1##
    C)##|f'(x)|>1##
    D)##|f'(x)|<1##

    2. Relevant equations
    MVT: $$f'(c)=\frac{f(b)-f(a)}{b-a}$$

    3. The attempt at a solution
    I first tried using integration.
    $$-1\le f''(x) \le 1$$
    integrating from 0 to x,
    $$-x\le f'(x)-f'(0) \le x$$
    Again integrating from 0 to x,
    $$-\frac{x^2}{2}\le f(x)-f(0)-f'(0)x \le \frac{x^2}{2}$$
    But even though I got an inequality for f(x), I could not remove the constants.
    Then I applied Rolle theorem for f(x). Since f(0)=f(1), there exists a point (at least one point) ##c## such that f'(c)=0.
    There exists a point ##X\in [c,x]## such that
    $$f''(X)=\frac{f(x)-f(c)}{x-c}$$
    Here, ##x\in [c,1]##, and since 1>x>c, x-c<1. Also, ##|f''(x)|\le 1##.
    $$f''(X)=\frac{f(x)-f(c)}{x-c}$$
    So, ##\frac{f(x)-f(c)}{x-c}\le 1## and ##f(x)-f(c)\le {x-c}##. Hence I get the answer D. Is this correct?
     
  2. jcsd
  3. Aug 14, 2015 #2

    pasmith

    User Avatar
    Homework Helper

    Metahints for real analysis:

    If you're given a continuous function on a closed bounded interval and told its values at the end points, consider applying the intermediate value theorem.
    If you're given a differentiable function on a closed bounded interval and told its values at the end points, consider applying the mean value theorem.

    There are two problems here. Firstly I take it you are applying the MVT to [itex]f'[/itex] rather than [itex]f[/itex], so [itex]X[/itex] satisfies [tex]
    f''(X) = \frac{f'(x) - f'(c)}{x - c} = \frac{f'(x)}{x - c}.[/tex]

    Secondly, writing [itex]X \in [c,x][/itex] implicitly requires that [itex]c \leq x[/itex]. But you have to prove a result for every [itex]x \in [0,1][/itex], so you have also to deal with the case [itex]x > c[/itex]. However that requires no further work, since as long as [itex]x \neq c[/itex] there is an [itex]X[/itex] lying between [itex]x[/itex] and [itex]c[/itex] which satisfies the above equation.

    Thus if [itex]x = c[/itex] then [itex]|f'(x)| = 0[/itex], and if [itex]x \neq c[/itex] then [itex]|f'(x)| < \dots[/itex]
     
  4. Aug 14, 2015 #3

    RUber

    User Avatar
    Homework Helper

    What does that inequality look like when x = 1? That will give you bounds on f'(0).
     
  5. Aug 14, 2015 #4

    RUber

    User Avatar
    Homework Helper

    What if you define ##f'(x) = f'(0) + \int_0^x f''(t) dt ##?
    Then using the inequality ## \left| \int f dx \right| \leq \int |f| dx##
    You can quickly deduce something about f'(x).

    Then, since f(0) = f(1), you can say that
    ##\int_0^c f'(x) dx = -\int_c^0 f'(x) dx##
    You can find a maximum for this based on the same inequality.

    I doubt you can say anything about |f(x)| since you aren't given anything that might constrain the initial values. f(0) = f(1) = 100 might be an option based on what you provided.
     
  6. Aug 15, 2015 #5

    Titan97

    User Avatar
    Gold Member

    @pasmith , that was a typo. I lost internet connection while trying to edit. @RUber, i found the minimum and maximum vaue f'(0) can take. $$-1/2 \le f'(0) \le 1/2$$
     
    Last edited: Aug 15, 2015
  7. Aug 15, 2015 #6

    Titan97

    User Avatar
    Gold Member

    But, from the equation, ##-x\le f'(x)-f(0)\le x## and by substituting the max value of f'(0), ##-x+0.5\le f'(x)\le x+0.5## and at x=1, f'(x)>1. Is that correct?
     
  8. Aug 15, 2015 #7
    Titan97 , do you want a proper solution , or would getting the answer alone simply be enough ?

    I haven't actually tried the question , but you could consider a function as f(x) = x(x - 1)/4 .

    Solving this can easily tell you about f'(x) ; however I don't believe you can comment on f(x) on the basis of what's given in the original question .

    I know this isn't a proper solution , but still ... Hope this helps .
     
  9. Aug 15, 2015 #8

    Titan97

    User Avatar
    Gold Member

    I want a proper solution. I know many substitutions that can give the answer to such questions.
     
  10. Aug 15, 2015 #9
    Okay , first of - You cannot comment on value of f(x) .

    For f'(x) - Let f'(x) = k ( f'(0) + x ) , where k is such that -1 <= k <= 1 .

    f'(0) = - 0.5 , and so f'(x) = k ( x - 0.5 ) .
    ( x - 0.5 ) varies from - 0.5 to 0.5 .
    ⇒ k ( x - 0.5 ) will belong to some interval lesser than equal ( - 0.5 , 0.5 ) - Depending on the value of k

    Hope this helps .
     
  11. Aug 15, 2015 #10

    Titan97

    User Avatar
    Gold Member

    what about post 6?
     
  12. Aug 15, 2015 #11
    Hint: There is a point [itex] x_0\in (0,1)[/itex] such that [itex]f'(x_0)=0[/itex]. If [itex]x[/itex] is another point in [itex](0,1)[/itex], then you can use the fact that [itex]|x-x_0|<1[/itex] together with the information about [itex]f''[/itex] to estimate [itex]f'(x)[/itex].
     
  13. Aug 15, 2015 #12
    F'(0) has a fixed value -
     
  14. Aug 15, 2015 #13
    Made a mistake here . I'll get back to you once I correct it .
     
  15. Aug 16, 2015 #14

    RUber

    User Avatar
    Homework Helper

    If you use the facts here, knowing that f'(c)=0, consider that f' goes directly from its initial value to zero as quickly as possible given constraints on f". The smallest c possible is zero. From there, assume max departure from zero using the max for f". How far from zero can f' get on the remainder of the unit interval?
     
  16. Aug 16, 2015 #15

    Titan97

    User Avatar
    Gold Member

    I did not understand that properly @RUber .
     
  17. Aug 16, 2015 #16
    I'm not sure why my post #11 is being ignored; it really suffices to solve the problem.
     
  18. Aug 16, 2015 #17

    Titan97

    User Avatar
    Gold Member

    Its the same thing i have given in the original post. It does work.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Question on Mean Value Theorem
Loading...