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Two Functions intersect and have equal derivatives at x=a and x=b

  1. Nov 12, 2013 #1
    Sorry, wasn't sure how to describe the problem in the title.

    The problem statement, all variables and given/known data

    Okay, this problem has really been bugging me for a while. It's a question that I thought of when I was daydreaming in class, but now I can't stop thinking about it. I've been going crazy for days Here is the problem:

    -

    Consider the following information to determine the answer to the question afterward.

    • f (0) and g (0) = 0
    • f '(0) and g '(0) = 1
    • f (h) and g (h) = 1
    • f '(h) and g '(h) = 0
    • h is positive constant greater than 1
    • f (x) and g (x) are always concave down on the interval (0, h)
    • f (x) and g (x) are both continuous on the interval (0, h)
    • f (x) and g (x) are both always differentiable on the interval (0, h)

    On the interval (0,h), do the graphs of f (x) and g (x) have to be identical?

    If not, give the algebraic expression of two functions that satisfy the above conditions and also have nonidentical graphs on the interval (0,h).

    The attempt at a solution

    I don't believe two functions necessarily have to be identical on the interval with those conditions. However, I can't find two functions to give an example

    My first attempt was to use f(x) = sinx as one of the functions, since it satisfies these conditions when h=pi/4.

    when f(x) = sinx, and h=pi/4
    • f (0) = 0
    • f '(0) = 1
    • f (pi/4) = 1
    • f '(pi/4) = 0

    Now I just needed to find a function g(x) with the same conditions, but has a curve nonidentical to sinx on the interval (0, pi/4). I figured if I manipulate the semi-circle equation, I can obtain a graph with the above characteristics. So I set g(x) = a√(b-cx^2) + d, which is a transformation of the semi-circle equation.

    However, after hours upon hours of attempts, I cannot find the values of a, b, c & d to satisfy all three conditions! For example, I can easily get g(0) = 0, g'(0) = 1, and g (pi/4) = 1, but then g '(pi/4) would equal something other than 0. What is happening?

    I also tried other functions like the cubic function, function raised to the 1/3 power, etc. but nothing works. There's always one condition unfulfilled.

    -

    Am I wrong in my thinking that two functions with the given conditions are not necessarily identical? I asked my GSI and he gave me some philosophical nonsense and told me I can figure it out on my own (he's obviously wrong).

    PS: Could you explain it in a way that a first-semester Calculus student can understand?
     
    Last edited: Nov 12, 2013
  2. jcsd
  3. Nov 12, 2013 #2

    jedishrfu

    Staff: Mentor

    Consider a sin curve vs a square wave with the same periodicity at specific points the value and slope might agree but the functions are clearly different.
     
  4. Nov 12, 2013 #3

    pasmith

    User Avatar
    Homework Helper

    Suppose there exists a function [itex]F: [0,h] \to \mathbb{R}[/itex], which satisfies the following conditions:
    • [itex]F[/itex] is continuous on [itex][0,h][/itex] and differentiable on [itex](0,h)[/itex].
    • [itex]F[/itex] is concave down on [itex](0,h)[/itex].
    • [itex]F(0) = F'(0) = F(h) = F'(h) = 0[/itex].
    • There exists [itex]x \in (0,h)[/itex] such that [itex]F(x) \neq 0[/itex].

    Now suppose [itex]f : [0,h] \to \mathbb{R}[/itex] satisfies the following:
    • [itex]f(0) = 0[/itex], [itex]f'(0) = 1[/itex], [itex]f(h) = 1[/itex] and [itex]f'(h) = 0[/itex].
    • [itex]f[/itex] is concave down on [itex](0,h)[/itex].
    • [itex]f[/itex] is continuous on [itex][0,h][/itex] and differentiable on [itex](0,h)[/itex].

    Then [itex]f[/itex] and [itex]g = f + F \neq f[/itex] will satisfy the conditions you have imposed.

    All you need do is find such an [itex]F[/itex].
     
    Last edited: Nov 12, 2013
  5. Nov 12, 2013 #4
    Yeah, that's why I don't think f(x) and g(x) have to be equal. But I can't think of any counterexamples to the proposed question. Can you?
     
  6. Nov 12, 2013 #5
    hmm, I'm a first-semester Calculus student. I assumed the question could be answered using some relatively basic calculus. I've never seen some of the symbols you just used.

    I doubt understand your third bullet where you put "F(0) = F'(0) = F(h) = F'(h) = 0." How did you figure that out? Also, for the fourth bullet, I have no idea what that "E" looking character is.

    Sorry for the hassle, but could you explain it in a way that a first-semester Calculus student can understand? I may be a few courses away from understanding your response completely.
     
  7. Nov 12, 2013 #6

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    @Jay520 re your OP: It's easy to draw a few such curves:

    curves.jpg
     
  8. Nov 12, 2013 #7

    pasmith

    User Avatar
    Homework Helper

    I want to show that if f and g satisfy the given conditions, then they need not be equal. Thus I want to write [itex]F = g - f[/itex], and show that [itex]F[/itex] doesn't have to be zero everywhere.

    Since [itex]f(0) = g(0)[/itex], [itex]f'(0) = g'(0)[/itex], [itex]f(h) = g(h)[/itex] and [itex]f'(h) = g'(h)[/itex] I therefore have [itex]F(0) = F'(0) = F(h) = F'(h) = 0[/itex].

    "[itex]a \in A[/itex]" means "[itex]a[/itex] is a member of the set [itex]A[/itex]".
     
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