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Continuity and Intermediate Value Theorem

  1. Sep 29, 2005 #1
    Hello everyone,

    I have come across two questions that I have solved, but unfortunately am quite sure I've done them incorrectly. They are related to continuity and the intermediate value theorem.

    Find the constant c that makes g continuous (-infinity,infinity).

    g(x){ x^2-c^2 if x<4
    { cx+20 if x>4

    For this question I found that the graphs is continuous from (-infinity,4),[4,infinity] Then I found the limits as x approaches 4 from the left and right. which ended up being 16-c^2 and 4c+20. I then made these expressions equal to each other to solve for the constant c and ended up getting c=-4 and c=8. Neither of these values work, and I'm not quite sure what I should have done.

    Use the I.V.T to show that there is a root of the given equation inthe specified interval
    tanx=2x (0,1.4)

    when f(0) you get 0
    When f(1.4) you get 2.99

    therefore f(0) <0<f(1.4)

    What I did here just seems wrong, there must be more to it than that, but that's all I can get from reading the textbook.

    Thanks again
  2. jcsd
  3. Sep 29, 2005 #2


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    Homework Helper

    1) You didn't solve for the constant correctly. There is only one possibility for c (i.e. if you're using the quadratic equation, the discriminant b² - 4ac = 0).

    2) First of all, 0 is not even in the interval (0, 1.4). And in fact, neither is 1.4. What you did is indeed wrong. Suppose I say we have the function f(x) = x. If your proof is valid, then there is a number c in that interval such that f(c) = 0 because f(0) = 0, f(1) = 1, and f(0) < 0 < f(1). What you need to do is show that there is a in (0, 1.4) such that f(a) < 0, and b in (0, 1.4) such that f(b) > 0. Given that f is continuous on all of the reals, the following would also suffice:

    f(0) < 0 and f(c) > 0 for some c in the interval
    f(0) > 0 and f(c) < 0 for some c in the interval
    f(1.4) < 0 and f(c) > 0 for some c in the interval
    f(1.4) > 0 and f(c) < 0 for some c in the interval

    The first three cannot be satisfied, but you can show the fourth. However, just to be safe (since you might get questions where f is only defined on the given interval) pick a and b INSIDE the interval (0, 1.4) such that f(a) < 0 and f(b) > 0. Better yet, find a such that f(a) < 0 and b such that f(b) > 0. This should be really easy, it's just a matter of picking numbers and you should be able to draw the graph of the function g(x) = tanx - 2x yourself and see where you can find numbers that do the trick. If you're allowed to use a calculator it should be even easier.
  4. Sep 30, 2005 #3
    Ok, I figured out the first one, I was just making a stupid mistake, thanks for you help on that. I'm still not sure of the last question, but I will keep at it.Thanks again.
  5. Sep 30, 2005 #4


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    Staff Emeritus
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    Knowing that f(0)= 0 doesn't help you because it is quite possible that f(x) might just rise from 0 up to f(1.4) without ever being equal to 0 again. And, as AKG pointed out, 0 is not in the interval and so does not count as a solution.

    Can you calculate f(1)= tan(1)- 2?
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