AP®︎/College Calculus AB
- Intermediate value theorem
- Worked example: using the intermediate value theorem
- Using the intermediate value theorem
- Justification with the intermediate value theorem: table
- Justification with the intermediate value theorem: equation
- Justification with the intermediate value theorem
- 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
that's continuous over the interval , the function will take any value between and over the interval.
More formally, it means that for any value
between and , there's a value in for which .
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
... then it must pass through any
-value between and .
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
with the following table of values. Let's find out where must there be a solution to the equation .
and . The function must take any value between and over the interval .
Which of the following is guaranteed by the Intermediate Value Theorem?
Want to try more problems like this? Check out this exercise.
Want to join the conversation?
- 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)
- 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(29 votes)
- 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)
- 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)
- How does the intermediate value theorem work?(0 votes)
- 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)
- 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)
- what if you're asked to use IVT to show that x^1/2 + (x+1)^1/2 =4 has a root(4 votes)
- 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.(6 votes)
- what the IVT,IRCand ARC stands for(1 vote)
- How to show the existence of a real number whose square is two using intermediate value theorem?(1 vote)
- The first question to ask is: What is the function? i.e. what is
The second question to ask is: What is a value for
xthat will make the function output a value less than 2? i.e. find a value for
f(a) < 2
The third question to ask is: What is a value for
xthat will make the function output a value greater than 2? i.e. find a value for
f(b) > 2
You then apply the intermediate value theorem to find that there must be a number between
bthat must give you 2 when squared.
Does that help?(6 votes)
- 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)
- 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)
- 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)
- 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)
- 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)
- 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)