Limits and Continuity – Intermediate Value Theorem (IVT)

Intermediate Value Theorem: show function has at least one solution — CALCULUS
Intermediate Value Theorem: show function has at least one solution — CALCULUS

Let’s say we have a function F(x) that is continuous on the interval [a, b]. The corresponding y values are F(a) and F(b), with F(a) being less than F(b).

Let a value M be in between F(a) and F(b).

The y value M corresponds with an x value C:

Of course, there may be more than one x value C that corresponds to the same y value M:

Therefore, the Intermediate Value Theorem is as follows:

Definition: Intermediate Value Theorem (IVT)

Let F(x) be a continuous function defined over the closed interval [a, b] such that

F(a) < F(b). Let m be a y value where F(a) < M < F(b), then there exists at least one value c where a < c < b, and that F(c) = M.

Of course, if F(b) < F(a) then there is some x value c such that a < c < b such that F(c) = M for F(b) < M < F(a).

Guidelines for applying the The Intermediate Value Theorem:

An application of the IVT is to verify that a function has at least one root (x-intercept). However, this would apply only if the signs of F(a) and F(b) are different, meaning that F(a) and F(b) are positive and negative, respectively (or negative and positive, respectively). This is called the Bozano’s Theorem, which states the following:

Definition: Bozano’s Theorem

If y = F(x) be a continuous function over the closed interval [a, b], then there has to be at least one x value c where a < c < b with F(c) = 0 if the signs of F(a) and F(b) are different.

Guidelines for applying Bozano’s Theorem:

Example 1: Does the function F(x) = x4-x3+3 on the interval of [1,2] have a zero?

Solution:

F(x) is a polynomial function, so we know that it will be continuous over the specified interval (or any interval for that matter).

We are given the interval [1,2], so a = 1, and b = 2.

First, let’s find F(a), which is f(1)

f(1) = (1)4-(1)3+3= 2.067

Next, let’s find F(b), which is f(2):

f(2) = (2)4-(2)3+3= -3.933

Since the signs of F(a) and F(b) are different, there has to be at least one zero of the function.

Let’s verify this by using a graphing calculator:

The root on the interval of [1,2] is x = 1.494.

Example 2: Does the function F(x) = cos(t)-3t on the interval of [0,1] have a zero?

Solution:

We are given the interval [0,1], so a = 0, and b = 1.

First, let’s find F(a), which is f(0)

f(0) = cos(0)-3(0) = 1

Next, let’s find F(b), which is f(1):

f(1) = cos(1)-3(1) = -2.46

Since the signs of F(a) and F(b) are different, there has to be at least one zero of the function.

Let’s verify this by using a graphing calculator:

Example 3: Use the Intermediate Value Theorem to show that F(x) = x2-x-12 has at least one zero on the interval of [2,5]. Use algebra to find the root(s) of the function within the specified interval.

Solution:

F(x) is a polynomial function, so we know that it will be continuous over the specified interval (or any interval for that matter).

We are given the interval [2,5], so a = 2, and b = 5.

First, let’s find F(a), which is f(2)

f(2) = (2)2-(2)-12= -10

Next, let’s find F(b), which is f(5):

f(5) = (5)2-(5)-12= 8

Since the signs of F(a) and F(b) are different, there has to be at least one zero of the function.

To find the value of the root(s) of the function, we will factor:

F(x) = x2-x-12

F(x) = (x+4)(x-3)

Set each factor equal to zero and solve for x:

x-4=0 —> x = 4

x-3=0 —> x = -3

Of course, you can always use the quadratic formula to find the roots and obtain the same answer:

x = -3 or x = 4

The roots of F(x) are x = -3 and x = 4. Now we need to pick the root(s) that are within the interval of [2,5]. x = 4 is the only root that satisfies this.

Let’s use a graphing calculator to verify the results:

Example 4: if F(x) = 2×2+3x+5, F(c) = 20, and the interval is [0,5], use the Intermediate Value Theorem to find the value(s) of c on the interval.

Solution:

First, let’s find f(0):

f(0) = 2(0)2+3(0)+5

f(0) = 5

Second, let’s find f(5):

f(5) = 2(5)2+3(5)+5

f(0) = 70

Since F(a) = 5 < F(c) = 20 < F(b)= 70, we can use the Intermediate Value Theorem.

We need the x value(s) that will make F(x) = 20. Therefore, set the function F(x) equal to 20:

20 = 2×2+3x+5

Subtract 20 from both sides:

0 = 2×2+3x-15

Let’s use the quadratic formula to find the roots of the function:

x = -3.589 or x = 2.089

The c values that will make F(x) = 20 are x = -3.589 and x = 2.089.

However, we need to find the x value that is on the interval of [0,5], and x = 2.089 is the only one that fits inside the interval.

Let’s find f(2.089) to verify our results:

f(2.089) = 2(2.089)2+3(2.089)+5= 20

Summary of Section

Intermediate Value Theorem: If a function is continuous on [a, b], and if M is any number between F(a) and F(b), then there must be a value, x = c, where a < c < b, such that F(c) = M.

Bozano’s Theorem: If a continuous function on a given defined interval changes sign, then it must equal zero at some point in the interval. There are two ways to find the solution to the function on the interval [a,b]:

References

https://www.mathsisfun.com/algebra/intermediate-value-theorem.html
https://www.khanacademy.org/math/ap-calculus-ab/ab-limits-new/ab-1-16/a/intermediate-value-theorem-review
https://www.slideshare.net/math265/45-continuous-functions-and-differentiable-functions
https://magoosh.com/hs/ap-calculus/2017/ap-calculus-review-intermediate-value-theorem/
https://study.com/academy/lesson/extreme-value-theorem-bolzanos-theorem.html
http://scidiv.bellevuecollege.edu/dh/Calculus_all/CC_1_3_ContinuousFcns

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