# 1.7 Intermediate Value Theoremap Calculus

Chapters

## Exercise 1

We cover all the topics in Calculus. I use the technique of learning by example. I Leave out the theory and all the wind. I work out examples because I know this is what the student wants to see. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Prove that the function f(x) = x²− 4x + 2 intersects the x-axis on the interval [0,2]. Can the same be said for the function: ?

## Exercise 2

Given the function:

Can it be said that f(x) is bounded in the interval [1,4]?

## Exercise 3

Given the function f(x) = x² + 1. Can it be said that the function exists for all values in the interval [1,5]?

## Exercise 4

Prove that the equation: x³+ x − 5 = 0, has at least one solution x = a such that 1 < a < 2.

## Exercise 5

Given the function f(x) = x³ − x² + 1. Can it be said that there is at least one point, c, inside the interval [1,2] which verifies that f(c) = 0?

## Exercise 6

Prove that the polynomial function f(x) = x³ + x + 1 has a value of zero between −1 and 0.

## Exercise 7

Prove that the equation e−x + 2 = x has at least one real solution.

## Exercise 8

Prove that there is a real number, x, such that sin x = x.

## Exercise 9

Given the function:

Prove that there is a point in the open interval (2, 4) in which the function f(x) has a value of 1.

## Exercise 10

Given the function f(x) = x³, determine if it is bounded superiorly and inferiorly in the interval [1, 5] and indicate if it reaches its maximum and minimum values within this interval.

## Exercise 11

Prove that the function f(x) = x + sin x − 1 is continuous at and prove that there exists at least one real root of the equation x + sin x − 1 = 0.

## Exercise 12

f and g are two continuous functions in [a, b] and such that f(a) > g(a) and f(b) < g(b). Prove the existance of c withinin (a, b) such that f(c) = g (c).

## Solution of exercise 1

Prove that the function f(x) = x²− 4x + 2 intersects the x-axis in the interval [0,2]. Can the same be said for the function: ?

The first function is continuous at .

f(0) = 0²− 4 · 0 + 2 > 0.

f(2) = 2²− 4 · 2 + 2 < 0.

Since it verifies the intermediate value theorem, there is at least one c that belongs to the interval (0, 2) and intersects the x-axis.

We cannot confirm the same of the second function because it is not continuous at x = 1.

## Solution of exercise 2

Given the function:

Can it be said that f(x) is bounded in the interval [1,4]?

Since f(x) is not continuous at x = 1, the function is not continuous in the closed interval [1,4], as a result it cannot be said that the function is bounded in that interval.

## Solution of exercise 3

Given the function f(x) = x² + 1. Can it be said that the function exists for all values in the interval [1,5]?

x² + 1 = 1 x = 0

x² + 1 = 5 x = 2

The function is continuous at R since it is a polynomial function.

It is in the interval [0, 2] as it is verified that f(0) = 1 and f(2) = 5.

Since it verifies the intermediate value theorem, the function exists at all values in the interval [1,5].

## Solution of exercise 4

Using Bolzano's theorem, show that the equation: x³+ x − 5 = 0, has at least one solution for x = a such that 1 < a < 2.

f(x) is continuous in [1,2]

f(1) = 1³+ 1 − 5 = −3 < 0

f(2) = 2³+ 2 − 5 = 5 > 0

Since it verifies the Bolzano's Theorem, there is c (1,2) such that:

f(c) = 0c³+ c − 5 = 0.

Therefore there is at least one real solution to the equation x³+ x − 5 = 0.

## Solution of exercise 5

Given the function f(x) = x³ − x² + 1. Can it be said that there is at least one point, c, inside the interval [1,2] which verifies that f(c) = 0?

f(x) is continuous in [1, 2].

f(1) = 1³ − 1² + 1 = 1 > 0.

f(2) = 2³ − 2² + 1 = 5 > 0.

The Bolzano theorem cannot be applied because it does not change sign.

## Solution of exercise 6

### 1.7 Intermediate Value Theoremap Calculus Solver

Prove that the polynomial function f(x) = x³ + x + 1 has a value of zero between −1 and 0.

f(x) is a polynomial and therefore is continuous in the interval [−1, 0].

f(−1) = (−1)³ + (−1) + 1 = −1 < 0.

f(0) = 0 + 0 + 1 = 1 > 0.

There is a c(−1, 0) such that f(c) = 0

## Solution of exercise 7

Prove that the equation e−x + 2 = x has at least one real solution.

The function is continuous in the interval [0, 3].

f(0) = e0 + 2 − 0 > 0.

f(3) = e—3 + 2 − 3 < 0.

Since it verifies Bolzano's theorem, there is c (0, 3) such that:

f(c) = 0 e−c + 2 = c.

Therefore there is at least one real solution to the equation e−x + 2 = x.

## Solution of exercise 8

Prove that there is a real number, x, such that sin x = x.

Consider the function f(x) = sin x − x.

It is continuous at .

f(−π) = sin (−π) − (−π) = 0 + π = π > 0

f(π) = sin (π) − (π) = 0 − π = −π < 0

There is a c (−π. π) such that:

f(c) = 0 sen c = c

Therefore, there is at least one real solution to the equation sin x = x.

## Solution of exercise 9

Given the function:

Prove that there is a point in the open interval (2, 4) in which the function f(x) has a value of 1.

The exponential function is positive at , therefore the denominator of the function cannot be annulled.

There is only doubt of the continuity at x = 0, which is out of the interval being studied, therefore f(x) is continuous in [2. 4].

Consider the function g defined by g(x) = f(x) − 1.

g is continuous on the interval [2. 4].

Since it verifies the intermediate value theorem, there is a c (2, 4) such that:

## Solution of exercise 10

Given the function f(x) = x³, determine if it is bounded superiorly and inferiorly in the interval [1, 5] and indicate if it reaches its maximum and minimum values within this interval:

The function is continuous in the interval [1, 5], as a result it can be affirmed that it is bounded in that interval.

Aswell as being continuous in the interval [1, 5], it has fulfilled the extreme value theorem, which affirms that it attains at least one maximum and absolute minimum in the interval [1, 5].

## Solution of exercise 11

Prove that the function f(x) = x + sin x − 1 is continuous at and prove that there exists at least one real root of the equation x + sin x − 1 = 0.

The function is continuous since it is the sum of continuous functions.

f(0) = 0 + sin 0 − 1 = − 1 < 0.

f(π/2) = π/2 + sin π/2 − 1 = π/2 > 0.

Since it verifies the intermediate value theorem, there is a c(o, π/2) such that:

f(c) = 0 c + sin c − 1 = 0

Therefore, there is at least one real solution to the equation x + sin x − 1 = 0.

## Solution of exercise 12

f and g are two continuous functions in [a, b] and such that f(a) > g(a) and f(b) < g(b). Compaq cq58 vga driver download 64-bit. Prove the existance of c withinin (a, b) such that f(c) = g (c).

h is the function defined by h(x) = f(x) − g(x).

Since f and g are continuous in [a, b], the function h also is.

f(a) > g(a) h(a) = f(a) − g(a) > 0

f(b) < g(b) h(b) = f(b) − g(b) < 0.

Since it verifies the intermediate value theorem, there is a c (a, b) such that:

h(c) = 0 f(c) − g(c) = 0 f(c) = g(c)

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In this section we learn a theoretically important existence theorem called theIntermediate Value Theorem and we investigate some applications.

### Intermediate Value Theorem

In this section we discuss an important theorem related to continuous functions.Before we present the theorem, lets consider two real life situations and observe animportant difference in their behavior. First, consider the ambient temperature andsecond, consider the amount of money in a bank account.

First, suppose that the temperature is at 8am and then suppose it is at noon.Because of the continuous nature of temperature variation, we can be surethat at some time between 8am and noon the temperature was exactly .Can we make a similar claim about money in a bank account? Supposethe account has \$65 in it at 8am and then it has \$75 in it at noon. Did ithave exactly \$70 in it at some time between 8 am and noon? We cannotanswer that question with any certainty from the given information. On onehand, it is possible that a \$10 deposit was made at 11am and so the total inthe bank would have jumped from \$65 dollars to \$75 without ever beingexactly \$70. On the other hand, it is possible that the \$10 was added in \$5increments. In this case, the account did have exactly \$70 in it at some time. Thefundamental reason why we can make certain conclusions in the first casebut cannot in the second, is that temperature varies continuously, whereasmoney in a bank account does not (it will have jump discontinuities). When aquantity is known to vary continuously, then if the quantity is observed to havedifferent values at different times then we can conclude that the quantitytook on any given value between these two at some time between our twoobservations. Mathematically, this property is stated in the Intermediate ValueTheorem.

Intermediate Value Theorem

If the function is continuous on the closed interval and is a number between and ,then the equation has a solution in the open interval .

The value in the theorem is called an intermediate value for the function on theinterval . Note that if a function is not continuous on an interval, then the equation may or may not have a solution on the interval.

Remark: saying that has a solution in is equivalent to saying that there exists anumber between and such that .

The following figure illustrates the IVT.

example 1 Show that the equation has a solution between and .
First, the function is continuous on the interval since is a polynomial. Second,observe that and so that 10 is an intermediate value, i.e., Now we can apply theIntermediate Value Theorem to conclude that the equation has a least one solutionbetween and . In this example, the number 10 is playing the role of in the statementof the theorem.
(problem 1) Determine whether the IVT can be used to show that the equation has asolution in the open interval .

### Intermediate Value Theorem Examples

Is continuous on the closed interval ?

and

Is an intermediate value?

YesNo

Can we apply the IVT to conclude that the equation has a solution in the openinterval ?

example 2 Show that the equation has a solution between and .

First, note that the function is continuous on the interval and hence it iscontinuous on the sub-interval, . Next, observe that and so that 2 is anintermediate value, i.e., Finally, by the Intermediate Value Theorem wecan conclude that the equation has a solution on the open interval . Inthis example, the number 2 is playing the role of in the statement of thetheorem.

(problem 2) Determine whether the IVT can be used to show that the equation has a solution in the open interval Is continuous on the closed interval ?

and

Is an intermendiate value?

YesNo

Does the IVT imply that the equation has a solution in the open interval ?

example 3 Show that the function has a root in the open interval .

Recall that a root occurs when . Since is a polynomial, it is continuous on theinterval . Plugging in the endpoints shows that 0 is an intermediate value: and so By the IVT, we can conclude that the equation has a solution (and hence has aroot) on the open interval .

(problem 3) Determine whether the IVT can be used to show that the function has aroot in the open interval .

Is continuous on the closed interval ?

and

Is an intermendiate value?

YesNo

### 1.7 Intermediate Value Theorem Ap Calculus Frq

Does the IVT imply that the function has a root in the open interval ?

example 4 Show that the equation has a solution between and .
Note that the equation is equivalent to the equation . The latter is the prefered formfor using the IVT. So let Since is the difference between two continuous functions, itis continuous on the closed interval . Next, we compute and and show that 0 is anintermediate value: and and so, By the IVT, the equation has a solution in theopen interval . Hence the equivalent equation has a solution on the sameinterval.
example 5 Use the IVT four times to approximate a root of the polynomial First, isa polynomial, so it is continuous on any closed interval. Next, note that By the IVT has a root in the interval . To use the IVT a second time, we now determine themidpoint of this interval: and we plug this in to : Combining this with we can usethe IVT again to conclude that has a root on the interval . This intervalis half of the original interval- the original interval has been bisected. Wenow use the IVT a third time. The midpoint of the interval is . We plugthis into : Combining this with we can use the IVT to conclude that hasa root on the interval . This interval is half of the previous interval- theprevious interval has been bisected. We will do this a fourth and final time.Note that is the midpoint of the interval and Combining this with wecan use the IVT a fourth time to conclude that has a root on the interval. At this stage, our approximation of the root is which is the midpointof this interval. Our error is then no more than , which is half the widthof the interval. We will stop here, but the method could theoretically becontinued indefinitely giving a better approximation to the root each time.This method of approximating roots is called the Method of ContinuedBisection.

### 1.7 Intermediate Value Theorem Ap Calculus Solver

Here is a detailed, lecture style video on the Intermediate Value Theorem: