If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Intermediate value theorem review

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 [a,b], the function will take any value between f(a) and f(b) over the interval.
More formally, it means that for any value L between f(a) and f(b), there's a value c in [a,b] for which f(c)=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 (a,f(a)) and (b,f(b))...
... then it must pass through any y-value between f(a) and f(b).
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(x)=2.
x2101
f(x)4311
Note that f(1)=3 and f(0)=1. The function must take any value between 1 and 3 over the interval [1,0].
2 is between 1 and 3, so there must be a value c in [1,0] for which f(c)=2.
Problem 1
f is a continuous function.
f(2)=3 and f(1)=6.
Which of the following is guaranteed by the Intermediate Value Theorem?
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)
    Default Khan Academy avatar avatar for user
  • 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)
    Default Khan Academy avatar avatar for user
    • 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.
      (26 votes)
  • duskpin ultimate style avatar for user Arjun Kavungal
    How does the intermediate value theorem work?
    (0 votes)
    Default Khan Academy avatar avatar for user
    • female robot grace style avatar for user elizabeth.s.gall
      The theorem basically says "If I pick an X value that is included on a continuous function, I will get a Y value, within a certain range, to go with it." We know this will work because a continuous function has a predictable Y value for every X value. By "predictable" I mean that the limits for a point from the right and left side are the same as the point's value in the function.

      A simple way to imagine this is to pretend the continuous function occupies a box. We don't necessarily know what the function looks like, but we know where it can and can't go. Let's say our interval is [0, 10] and the end points are (0, 1) and (10, 5). In that case we know the function can't go any further left than 0, or any further right than 10. On the y-axis, our function could go lower than 1 or higher than 5 - we don't know. But the important thing is, it definitely has to go between one and five, otherwise it wouldn't connect our end points!

      In answering a multiple-choice question about values given by the Intermediate Value Theorem, imagine this box on your graph. If the any option makes assumptions about what happens outside that box, don't select it, because there's no way for you to know those things from the information you've been given.
      (31 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
    (4 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Angel
      We have f(x) = x^(1/2) + (x+1)^(1/2) - 4

      We need to show some closed interval [a, b] where f(x) is continuous.
      We also need to show that 0 is between f(a) and f(b)

      First lets establish a closed interval where the function is continuous.

      f(x) is continuous for x >= 0 since the function is made by adding multiple square root functions which are also continuous for x>= 0.

      Second, lets find a, and b by experimenting with different x-values.

      f(0) = 0^(1/2) + (0+1)^(1/2) - 4
      f(0) = 0 + 1 - 4
      f(0) = -3

      f(5) = 5^(1/2) + (5+1)^(1/2) - 4
      f(5) is approximately 0.6856

      Notice that 0 is between f(0) and f(5).

      Finally, lets summarize the information we found.

      We know that f(x) is continuous for x >= 0.
      This means that f(x) is also continuous across [0, 5].
      We found that f(0) = -3 and f(5) is approximately 0.6856
      We now know that 0 is between f(0) and f(5).

      Thus, f(x) has a root somewhere in the interval 0 <= x <= 5.

      You could make a similar argument with other intervals that you find.
      (7 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)
    Default Khan Academy avatar avatar for user
  • aqualine seed style avatar for user Mariem Bakr
    what the IVT,IRCand ARC stands for
    (1 vote)
    Default Khan Academy avatar avatar for user
  • 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?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • 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 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)
    Default Khan Academy avatar avatar for user
    • 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)
  • 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)
    Default Khan Academy avatar avatar for user
    • 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)
  • 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)
    Default Khan Academy avatar avatar for user
    • 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)