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Intermediate Value Theorem

Q: What is the intermediate value theorem?

The Intermediate Value Theorem says that if a function has no discontinuities, then there is a point which lies between the endpoints whose y-value is between the y-values of the endpoints.

Q: What is the Intermediate Value Theorem formula?

The Intermediate Value Theorem guarantees that if a function f is continuous on the interval [ a , b ] and has a function value N such that f(a) < N < f(b ) where f(a) and f(b) are not equal, then there is at least one number c in ( a , b ) such that f(c) = N .

Q: What is the Intermediate Value Theorem and why is it important?

The Intermediate Value Theorem says that if a function has no discontinuities, then there is a point which lies between the endpoints whose y-value is between the y-values of the endpoints. The IVT is a foundational theorem in Mathematics and is used to prove numerous other theorems, especially in Calculus.

Q: How do you prove the intermediate value theorem?

To prove the Intermediate Value Theorem, ensure that the function meets the requirements of the IVT. In other words, check if the function is continuous and check that the target function value lies between the function value of the endpoints. Then and only then can you use the IVT to prove a solution exists.

Q: How to use the Intermediate value theorem?

To use the Intermediate Value Theorem: First define the function f(x) Find the function value at f(c) Ensure that f(x) meets the requirements of IVT by checking that f(c) lies between the function value of the endpoints f(a) and f(b) Lastly, apply the IVT which says that there exists a solution to the function f

Imagine you take off on an airplane at 100 meters above sea level. The plane climbs very quickly, reaching an altitude of 1000 meters 5 minutes later. It would be safe to say that between the time you took off and the time you reached 1000 meters, there must have been a point where you attained an altitude of 500 meters, right? This may seem to be a trivial concept, but a very important one in Calculus! This concept stems from the Intermediate Value Theorem (IVT).

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If a function meets the Intermediate Value Theorem requirements, ______ solution is/are guaranteed.

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Last Updated: 04.07.2022
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The IVT answers a crucial question in Mathematics: does an equation have a solution? This article will define the Intermediate Value Theorem, discuss some of its uses and applications, and work through examples.

Intermediate Value Theorem Definition

The Intermediate Value Theorem states that if a function $f$ is continuous on the interval $[a, b]$ and a function value N such that $f (a) < N < f (b)$ where $f (a) \neq f (b)$ , then there is at least one number $c$ in $(a, b)$ such that $f (c) = N$ .

Essentially, IVT says that if a function has no discontinuities, there is a point between the endpoints whose y-value is between the y-values of the endpoints. The IVT holds that a continuous function takes on all values between $f (a)$ and $f (b)$ .

Intermediate Value Theorem geometric representation Vaia Since the function is continuous, IVT says that there is at least one point between a and b that has a y-value between the y-values of a and b - Vaia Original

Uses and Applications of the Intermediate Value Theorem in Calculus

The Intermediate Value Theorem is an excellent method for solving equations. Suppose we have an equation and its respective graph (pictured below). Let's say we are looking for a solution to $c$ . The Intermediate Value Theorem says that if the function is continuous on the interval $[a, b]$ and if the target value that we're searching for is between $f (a)$ and $f (b)$ , we can find $c$ using $f (c)$ .

Intermediate Value Theorem graphical representation existence of c Vaia The Intermediate Value Theorem guarantees the existence of a solution c - Vaia Original

The Intermediate Value Theorem is also foundational in the field of Calculus. It is used to prove many other Calculus theorems, namely the Extreme Value Theorem and the Mean Value Theorem.

Examples of the Intermediate Value Theorem

Example 1

Prove that $x^{3} + x - 4 = 0$ has at least one solution. Then find the solution.

Step 1: Define f(x) and graph

We'll let $f (x) = x^{3} + x - 4$

Intermediate Value Theorem example graph Vaia

Step 2: Define a y-value for c

From the graph and the equation, we can see that the function value at $c$ is 0.

Step 3: Ensure f(x) meets the requirements of the IVT

From the graph and with a knowledge of the nature of polynomial functions, we can confidently say that $f (x)$ is continuous on any interval we choose.

We can see that the root of $f (x)$ lies between 1 and 1.5. So, we'll let our interval be [1, 1.5]. The Intermediate Value Theorem says that $f (c) = 0$ must lie between $f (a)$ and $f (b)$ . So, we plug in and evaluate $f (1)$ and $f (1.5)$ .

$\begin{array}{rcl} f (1) & < & f (c) < f (1.5) \\ 1^{3} + 1 - 4 & < & 0 < 1 . 5^{3} + 1.5 - 4 \\ - 2 & < & 0 < 0.875 \end{array}$

Step 4: Apply the IVT

Now that all of the IVT requirements are met, we can conclude that there is a value $c$ in $[1, 1.5]$ such that $f (c) = 0$ .

So, $f (x)$ is solvable.

Example 2

Does the function $f (x) = x^{2}$ take on the value $f (x) = 7$ on the interval $[1, 4]$ ?

Step 1: Ensure f(x) is continuous

Next, we check to make sure the function fits the requirements of the Intermediate Value Theorem.

We know that $f (x)$ is continuous over the entire interval because it is a polynomial function.

Step 2: Find the function value at the endpoints of the interval

Plugging in $x = 1$ and $x = 4$ to $f (x)$

$f (1) = 1^{2} = 1 f (4) = 4^{2} = 16^{}$

Step 3: Apply the Intermediate Value Theorem

Obviously, $1 < 7 < 16$ . So we can apply the IVT.

Now that all IVT requirements are met, we can conclude that there is a value $c$ in [1, 4] such that $f (c) = 7$ .

Thus, $f (x)$ must take on the value 7 at least once somewhere in the interval $[1, 4]$ .

Remember, the IVT guarantees at least one solution. However, there may be more than one!

Example 3

Prove the equation $\frac{x - 1}{x^{2} + 2} = \frac{3 - x}{1 + x}$ has at least one solution on the interval $[- 1, 3]$ .

Let's try this one without using a graph.

Step 1: Define f(x)

To define $f (x)$ , we'll factor the initial equation.

$\begin{array}{rcl} (x - 1) (x + 1) & = & (3 - x) (x^{2} + 2) \\ x^{2} - 1 & = & - x^{3} + 3 x^{2} - 2 x + 6 \\ x^{3} - 2 x^{2} + 2 x - 7 & = & 0 \end{array}$

So, we'll let $f (x) = x^{3} - 2 x^{2} + 2 x - 7$

Step 2: Define a y-value for c

From our definition of $f (x)$ in step 1, $f (c) = 0$ .

Step 3: Ensure f(x) meets the requirements of the IVT

From our knowledge of polynomial functions, we know that $f (x)$ is continuous everywhere.

We will test our interval bounds, making $a = - 1 a n d b = 3$ . Remember, using the IVT, we need to confirm

$f (a) < f (c) < f (b) f (a) < 0 < f (b)$

Let $a = - 1$ :

$\begin{array}{rcl} f (a) & = & f (- 1) \\ = & {(- 1)}^{3} - 2 {(- 1)}^{2} + 2 (- 1) - 7 \\ = & - 12 \end{array}$

Let b= 3:

$\begin{array}{rcl} f (b) & = & f (3) \\ = & 3^{3} - 2 {(3)}^{2} + 2 (3) - 7 \\ = & 8 \end{array}$

Therefore, we have

$f (a) < f (c) < f (b) - 12 < 0 < 8$

Therefore, but the IVT, we can guarantee there is at least one solution to

$x^{3} - 2 x^{2} + 2 x - 7 = 0$

on the interval $[- 1, 3]$ .

Step 4: Apply the IVT

Now that all IVT requirements are met, we can conclude that there is a value $c$ in [0, 3] such that $f (c) = 0$ .

So, $f (x)$ is solvable.

Proof of the Intermediate Value Theorem

To prove the Intermediate Value Theorem, grab a piece of paper and a pen. Let the left side of your paper represent the y-axis, and the bottom of your paper represent the x-axis. Then, draw two points. One point should be on the left side of the paper (a small x-value), and one point should be on the right side (a large x-value). Draw the points such that one point is closer to the top of the paper (a large y-value) and the other is closer to the bottom (a small y-value).

The Intermediate Value Theorem states that if a function is continuous and if endpoints $a$ and $b$ exist such that $f (a) \neq f (b)$ , then there is a point between the endpoints where the function takes on a function value between $f (a)$ and $f (b)$ . So, the IVT says that no matter how we draw the curve between the two points on our paper, it will go through some y-value between the two points.

Try to draw a line or curve between the two points (without lifting your pen to simulate a continuous function) on your paper that does not go through some point in the middle of the paper. It is impossible, right? No matter how you draw a curve, it will go through the middle of the paper at some point. So, the Intermediate Value Theorem holds.

Intermediate Value Theorem - Key takeaways

The Intermediate Value Theorem states that if a function f is continuous on the interval [a, b] and a function value N such that $f (a) < N < f (b)$ where $f (a) \neq f (b)$ , then there is at least one number $c$ in $(a, b)$ such that $f (c) = N$
- Essentially, the IVT holds that a continuous function takes on all values between $f (a)$ and $f (b)$
IVT is used to guarantee a solution/solve equations and is a foundational theorem in Mathematics
To prove that a function has a solution, follow the following procedure:
- Step 1: Define the function
- Step 2: Find the function value at $f (c)$
- Step 3: Ensure that $f (x)$ meets the requirements of IVT by checking that $f (c)$ lies between the function value of the endpoints $f (a)$ and $f (b)$
- Step 4: Apply the IVT

Flashcards in Intermediate Value Theorem

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If a function meets the Intermediate Value Theorem requirements, ______ solution is/are guaranteed.

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Frequently Asked Questions about Intermediate Value Theorem

What is the intermediate value theorem?

The Intermediate Value Theorem says that if a function has no discontinuities, then there is a point which lies between the endpoints whose y-value is between the y-values of the endpoints.

What is the Intermediate Value Theorem formula?

The Intermediate Value Theorem guarantees that if a function f is continuous on the interval [a, b] and has a function value N such that f(a) < N < f(b) where f(a) and f(b) are not equal, then there is at least one number c in (a, b) such that f(c) = N.

What is the Intermediate Value Theorem and why is it important?

The Intermediate Value Theorem says that if a function has no discontinuities, then there is a point which lies between the endpoints whose y-value is between the y-values of the endpoints. The IVT is a foundational theorem in Mathematics and is used to prove numerous other theorems, especially in Calculus.

How do you prove the intermediate value theorem?

To prove the Intermediate Value Theorem, ensure that the function meets the requirements of the IVT. In other words, check if the function is continuous and check that the target function value lies between the function value of the endpoints. Then and only then can you use the IVT to prove a solution exists.

How to use the Intermediate value theorem?

To use the Intermediate Value Theorem:

First define the function f(x)
Find the function value at f(c)
Ensure that f(x) meets the requirements of IVT by checking that f(c) lies between the function value of the endpoints f(a) and f(b)
Lastly, apply the IVT which says that there exists a solution to the function f

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