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Intermediate value theorem review

AP.CALC:
FUN‑1 (EU)
,
FUN‑1.A (LO)
,
FUN‑1.A.1 (EK)
Review the intermediate value theorem and use it to solve problems.

What is the intermediate value theorem?

The intermediate value theorem describes a key property of continuous functions: for any function f that's continuous over the interval open bracket, a, comma, b, close bracket, the function will take any value between f, left parenthesis, a, right parenthesis and f, left parenthesis, b, right parenthesis over the interval.
More formally, it means that for any value L between f, left parenthesis, a, right parenthesis and f, left parenthesis, b, right parenthesis, there's a value c in open bracket, a, comma, b, close bracket for which f, left parenthesis, c, right parenthesis, equals, L.
This theorem makes a lot of sense when considering the fact that the graphs of continuous functions are drawn without lifting the pencil. If we know the graph passes through left parenthesis, a, comma, f, left parenthesis, a, right parenthesis, right parenthesis and left parenthesis, b, comma, f, left parenthesis, b, right parenthesis, right parenthesis...
... then it must pass through any y-value between f, left parenthesis, a, right parenthesis and f, left parenthesis, b, right parenthesis.
Want to learn more about the intermediate value theorem? Check out this video.

What problems can I solve with the intermediate value theorem?

Consider the continuous function f with the following table of values. Let's find out where must there be a solution to the equation f, left parenthesis, x, right parenthesis, equals, 2.
xminus, 2minus, 101
f, left parenthesis, x, right parenthesis43minus, 11
Note that f, left parenthesis, minus, 1, right parenthesis, equals, 3 and f, left parenthesis, 0, right parenthesis, equals, minus, 1. The function must take any value between minus, 1 and 3 over the interval open bracket, minus, 1, comma, 0, close bracket.
2 is between minus, 1 and 3, so there must be a value c in open bracket, minus, 1, comma, 0, close bracket for which f, left parenthesis, c, right parenthesis, equals, 2.
Problem 1
f is a continuous function.
f, left parenthesis, minus, 2, right parenthesis, equals, 3 and f, left parenthesis, 1, right parenthesis, equals, 6.
Which of the following is guaranteed by the Intermediate Value Theorem?
Choose 1 answer:
Choose 1 answer:

Want to try more problems like this? Check out this exercise.

Want to join the conversation?

  • purple pi purple style avatar for user Cassidy Graham
    If an equation of a cube root function is given and you are asked to find an interval that has least one solution, how would you go about that. I understand the Intermediate Value Theorem, but I'm not sure how to find the the interval given an equation.
    (18 votes)
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    • leaf green style avatar for user kubleeka
      You find an x-coordinate where you know the function is negative and another where you know the function is positive. Then the interval with those endpoints must contain a solution, by Intermediate Value Theorem
      (27 votes)
  • blobby green style avatar for user Sofia L
    What if you're not given an interval to test within?

    For example: Use the Intermediate Value Theorem to show that the equation x^3 + x + 1 = 0 has at least 1 solution.
    (6 votes)
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    • leaf green style avatar for user kubleeka
      Then find an interval. By just picking x-values, we can see that your polynomial is positive at x=1 and negative at x= -1. So it must have a solution in (-1, 1). If you want a smaller interval than that, you can check the value at the halfway point. This will either force your zero into one half of (-1, 1) or the other, or it will actually find your zero. Once you have your new, smaller interval, you can continue to shrink it as much as you like.
      (3 votes)
  • blobby green style avatar for user Lukas Mitchell
    How would you solve this with a step by step answer:

    Suppose that f is a continuous function on the interval [0,1] such that 0 smaller than or equal to f(x) is greater than or equal to 1 for each x in [0,1]. Show that there is a number c in [0,1] such f(c)=c
    (2 votes)
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    • primosaur seed style avatar for user Ian Pulizzotto
      I assume you mean 0 smaller than or equal to f(x) is smaller than or equal to 1 for each x in [0,1].

      Define the function g(x) = f(x) - x. Because x is a continuous function, f(x) is a continuous function, and the difference of two continuous functions is continuous, g(x) is continuous.

      Note that g(0) = f(0) and g(1) = f(1) - 1.

      Since 0 <= f(x) <= 1 for each x in [0, 1], g(0) >= 0 and g(1) <= 0. Therefore, because g(x) is continuous, it follows from the intermediate value theorem that g(c) = 0 for some number c in [0,1]. Because g(c) = f(c) - c, f(c) = c for this number c.
      (3 votes)
  • duskpin tree style avatar for user k8
    One question asks "Why doesn't the IVT apply to g(x) on the interval [0, 4]. g(x)=2x^2-8x-5.

    It's continuous, so why doesn't it apply?
    (2 votes)
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  • starky tree style avatar for user nishvb
    I understand that the intermediate value theorem lets you confirm that a solution for f(c) exists within a certain interval, but what would you do for a problem that asks you to approximate a certain value (in my case, a cubic function's roots) using the intermediate value theorem?
    (1 vote)
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    • mr pink red style avatar for user andrewp18
      Well first I would find an interval [𝑎, 𝑏] where 𝑓 is monotonically increasing or decreasing, such that 𝑓(𝑎) < 0 < 𝑓(𝑏). Then by the Intermediate value theorem, there exists a 𝑐 ∈ (𝑎, 𝑏) such that 𝑓(𝑐) = 0, that is, 𝑐 is a root of 𝑓. If you were to strictly only use IVT, the best you can do here is bringing 𝑎 closer and closer to 𝑏 and testing the values of 𝑓 that result until you find a value in the interval that is sufficiently close to 𝑐 (or you could start from 𝑏 and use smaller and smaller values). This is pretty tedious, so I would rather use Newton's method to approximate the roots within this interval. Essentially, I take the local linearization at a point in the interval, then I find the 𝑥-intercept of that tangent line. I then repeat the process at the point on the curve which is the 𝑥-intercept of the first linearization. I'll get closer and closer to the root as a perform more and more steps. There are some certain circumstances where Newton's method will actually fail to find a root. You can learn more about local linearization and Newton's method on KA.
      Comment if you have questions!
      (4 votes)
  • aqualine seed style avatar for user Mariem Bakr
    what the IVT,IRCand ARC stands for
    (1 vote)
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  • blobby green style avatar for user Yashveer Man
    How to show the existence of a real number whose square is two using intermediate value theorem?
    (0 votes)
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    • female robot grace style avatar for user tyersome
      The first question to ask is: What is the function? i.e. what is f(x)

      The second question to ask is: What is a value for x that will make the function output a value less than 2? i.e. find a value for a so that f(a) < 2

      The third question to ask is: What is a value for x that will make the function output a value greater than 2? i.e. find a value for b so that f(b) > 2

      You then apply the intermediate value theorem to find that there must be a number between a and b that must give you 2 when squared.

      Does that help?
      (6 votes)
  • blobby green style avatar for user Praise Ogujiofor
    what if you're asked to use IVT to show that x^1/2 + (x+1)^1/2 =4 has a root
    (2 votes)
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  • starky seedling style avatar for user Selenashen
    How do you proof this by using the IVT?. Given any triangle ABC, prove that it is always possible to inscribe a square so that two vertices of the square lie on AB of the triangle, and the other two on
    AC and BC respectively.
    (1 vote)
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    • leaf green style avatar for user kubleeka
      Show that there exists a square with two vertices on AB, one on AC, and one inside the triangle. Show that there exists a second square with vertices on the same sides, and one vertex outside the triangle.

      Then the square you're looking for exists by the Intermediate Value Theorem.
      (2 votes)
  • blobby green style avatar for user halie jones
    When is the Intermediate Value Theorem DNE?
    (1 vote)
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