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: Functions, Continuity

  1. Sep 23, 2005 #1
    I recently finished a homework assignment with the exceptions of the following:

    1.) f(x) =x^3 - x^2 + x, show there is a number c such that f(c)=10.

    f(x) can be equated to 10, but I'm not quite sure how to solve the equation from that point.

    2.) Prove that the equation has at least one real root.

    e^x = 2 - x

    In order to understand this question, i attempted to carry the same procedure with another equation: y = x^2 + x + 2. If the discriminant is 0, then there is a single root. If the discriminant is <0, no roots, and >0, multiple roots. But the same procedure doesn't work with the above equation, or for cubics, quartics, etc.

    3.) For what values of x is F continuous?

    f(x) = [ 0 if x is rational, 1 if x is irrational

    I understand that the function can never be continuous, since it oscillates between 0 and 1 infinitely. But can anyone clarify what the following text means:
    http://mathworld.wolfram.com/images/equations/DirichletFunction/equation3.gif [Broken]
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Sep 23, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    1) You want to show that there is some x solving the equation:

    x³ - x² + x - 10 = 0

    Well you know the polynomial p defined by p(x) = x³ - x² + x - 10 is continuous, so you can apply the intermediate value theorem. That is, if you can find a number x' such that p(x') < 0 and a number x'' such that p(x'') > 0 then you know that there is a number c between x' and x'' (x' may be larger than x'' or may be smaller, doesn't matter) such that p(c) = 0, and such a c will be the c you're looking for.

    2) Try same technique as above.

    3) You're right that f is not continuous, but are you sure you know why? The function that is 0 on (0,1) then 1 on (1,2) then 0 again on (2,3) then 1 on (3, 4) then 0 on (4,5) then 1 on (5,6), etc. also "oscillates" between 0 and 1, but it has many points of continuity. Also, I don't know if "oscillates" is the correct term. Chances are you know why f is nowhere continuous, and your justification, "since it oscillates between 0 and 1 infinitely" was just an abbreviated version of the actual reasoning, but just in case it's not and you think that statement is sufficient, you should probably do a [itex]\epsilon - \delta[/itex] proof to be sure you know why. Either way, it would be good exercise to do the proof.

    What is it that you don't understand about the function [itex]D_M[/itex]? It is 0 when x is irrational, and 1/b when x is rational. But what's b? Well when x is rational, it can be expressed as the ratio of two integers, i.e. as a/b. For each rational you know there is a reduced form, so 2/4, 100/200, and 1/2 are all the same number but 1/2 is the reduced form. In this case, b=2, so f(100/200) = f(0.5) = f(1/2) = 1/2. Another example, f(100/150) = 1/3. This function is continuous at every irrational and discontinuous at every rational. Can you prove it?
  4. Sep 24, 2005 #3
    Here is how to solve a 3rd degree equation;

    General cubic form:
    y^3 + by^2 + cy + d=0

    1.Let y=x-b/3

    show that;

    x^3 + px + q=0

    2.In the equation
    x^3 + px + q=0

    Let x=H+K, such as 3HK=-p

    3.Show that

    -q=H^3 + K^3

    4.Solve for K in 3HK=-p and replace in H^3 + K^3=-q then show that

    H=(-q/2 +/- (q^2/4 + p^3/27)^1/2)^1/3

    5. Repeating step 3 and 4 we find out that

    K=(-q/2 +/- (q^2/4 + p^3/27)^1/2)^1/3 also.

    6.Since x=H+K, then

    x=(-q/2 +/- (q^2/4 + p^3/27)^1/2)^1/3 + (-q/2 +/- (q^2/4 +p^3/27)^1/2)^1/3

    Then solve for y.
  5. Sep 24, 2005 #4


    User Avatar
    Science Advisor

    Oh, please don't! The point of this problem was NOT to solve the equation but to use continuity to show that it HAS a solution.
    f(x)= x3 - x^2 + x

    f(0)= 0, f(1)= 1- 1+1= 2, f(2)= 8- 4+ 2= 6, f(3)= 27- 9+ 3= 21!

    I now know, because f(x) is a polynomial and so continuous for all x, that there exist a number x (in fact, between 2 and 3) such that f(x)= 10.
  6. Sep 24, 2005 #5
    You could also use the odd degree theorem;

    x^3 - x^2 + x -10, has to have a solution c such as f(c)=0

    Proof: For a large positive value of x^3, x^3> x^2 + 10 - x
    then f(x)>0
    For a small negative value of x, x^3 <x^2 + 10 - x
    then f(x)<0

    Since the function is continuous, it has to have a 0.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook