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The Intermediate Value Theorem We already know from the definition of continuity at a point that the graph of a function will not have a hole at any point where it is continuous. The Intermediate Value Theorem basically says that the. In fact, the intermediate value theorem is equivalent to the least upper bound property. Suppose the intermediate value theorem holds, and for a nonempty set S S S with an upper bound, consider the function f f f that takes the value 1 1 1 on all upper bounds of S S S and − 1-1 − 1 on the rest of R.

Chapters

Exercise 1

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:

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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

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. Wd my passport for mac manual.

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

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f(π) = sin (π) − (π) = 0 − π = −π < 0

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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). 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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The Intermediate Value Theorem

Your teacher probably told you that you can draw the graph of acontinuous function without lifting your pencil off the paper.This is made precise by the following result:

Intermediate Value Theorem. Let f (x) be a continuousfunction on the interval [a, b]. If d [f (a), f (b)], thenthere is a c [a, b] such that f (c) = d.

In the case where f (a) > f (b), [f (a), f (b)] is meant to be thesame as [f (b), f (a)]. Another way to state the IntermediateValue Theorem is to say that the image of a closed interval undera continuous function is a closed interval. We will present anoutline of the proof of the Intermediate Value Theorem on thenext page.

Here is a classical consequence of the Intermediate Value Theorem:

Example. Every polynomial of odd degree has at least onereal root.

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We want to show that if P(x) = a_nxⁿ + a_{n - 1}x^{n - 1} + ^.. + a₁x + a₀ is a polynomial with n odd and a_n 0,then there is a real number c, such that P(c) = 0.

First let me remind you that it follows from the results inprevious pages that every polynomial is continuous on the realline. There you also learned that

Consequently for | x| large enough, P(x) and a_nxⁿ havethe same sign. But a_nxⁿ has opposite signs for positive xand negative x. Thus it follows that if a_n > 0, there are realnumbers x₀ < x₁ such that P(x₀) < 0 and P(x₁) > 0. Similarlyif a_n < 0, we can find x₀ < x₁ such that P(x₀) > 0 andP(x₁) < 0. In either case, it now follows directly from theIntermediate Value Theorem that (for d = 0) there is a realnumber c [x₀, x₁] with P(c) = 0.

The natural question arises whether every function whichsatisfies the conclusion of the Intermediate Value Theorem mustbe continuous. Unfortunately, the answer is no andcounterexamples are quite messy. The easiest counterexample isthe function

As you found out on an earlier page, this function fails to becontinuous at x = 0. On the other hand, it is not too hard to seethat f

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(x) has the 'Intermediate Value Property' even on closedintervals containing x = 0.

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