# Calculus Of One Real Variable – By Pheng Kim Ving Chapter 1: Limits And Continuity – Section 1.2.3: The Intermediate-Value Theorem

1.2.3
The Intermediate-Value Theorem

# 1. The Intermediate-Value Theorem

Suppose f is a continuous function on [a, b]. See Fig. 1.1. Let v be a real number between f(a) and f(b). Intuitively,
since  f is continuous, it takes on every number between f(a) and f(b), ie, every intermediate value. Thus, v is a value of
f, which means that there exists c between a and b such that f(c) = v. ### Fig. 1.1

f(c) = v.

Theorem 1.1 – The Intermediate-Value Theorem

 If f is continuous on [a, b] and v lies between f(a) and f(b), then there exists c between a and b such that f(c) = v.

The proof of this theorem needs the following principle.

Dedekind Cut Principle

Let's partition [a, b] into two sets A and B such that:
i.   Each number in [a, b] is in either A or B but not both, and
ii.  Each number in A is less than every number in B.
See Fig. 1.2. ### Fig. 1.2

Dedekind Cut Partition Of [a, b]. Proof Of The Intermediate-Value Theorem   ### Fig. 1.3 EOP # 2. Applications Of The Intermediate-Value Theorem

#### Example 2.1

Prove that the cubic equation x3 + x2 – 4 = 0 has a solition in the interval (1, 2).

# Note

A graph of y = x3 + x2 – 4 is sketched in Fig. 2.1. We have:  ### Fig. 2.1

Graph Of y = x3 + x2 – 4.

Solution
Let f(x) = x3 + x2 – 4. Since f is a polynomial, it's continuous on [1, 2]. We have f(1) = 13 + 12 – 4 = –2 < 0 and f(2) =
23 + 22 – 4 = 8 > 0, so that f(1) < 0 < f(2). So by the intermediate-value theorem there exists x1 in (1, 2) such that f(x1)
= 0. That is, the equation x3 + x2 – 4 = 0 has a solution in the interval (1, 2).

EOS

Remark that in the solution we don't have to sketch a graph if not asked to.

#### Example 2.2

Determine the intervals where f(x) = x3 – 9x is positive and where it's negative.

# Solution EOS

Since f has a constant sign on each of the intervals, to determine its sign on an interval we evaluate its value at a point
inside that interval. The sign of that value is the sign of f on that interval. The point chosen inside each interval should be
the one that makes the calculation of the value of f at it as simple as possible.

## Problems & Solutions

1.  Show that the equation x3 + x – 1 = 0 has a solution in the interval (0, 1).

#### Solution

Let f(x) = x3 + x – 1. We have f(0) = 03 + 0 – 1 = –1 < 0 and f(1) = 13 + 1 – 1 = 1 > 0, so that f(0) < 0 < f(1). Thus by
the intermediate-value theorem there exists c in (0, 1) such that f(c) = 0. That is, the equation x3 + x – 1 = 0 has a
solution in the interval (0, 1). 2.  Prove that the function f defined by f(x) = x3 15x + 1 has at least three zeros in [– 4, 4]. (A point x1 is a zero of f if
f(x1) = 0.)

Solution We have: f(– 4) = –3 < 0, f(0) = 1 > 0, f(1) = –13 < 0, and f(4) = 5 > 0. Since f is continuous on [– 4, 0] and f(– 4) < 0
< f(0), the Intermediate-Value Theorem assures us that there exists at least one number a in (– 4, 0) such that f(a) = 0,
ie, f has at least one zero in (– 4, 0). Similarly, f has at least one zero in (0, 1) and at least one zero in (1, 4). Hence, f
has at least three zeros in (– 4, 0) U (0, 1) U (1, 4), thus at least three zeros in [– 4, 4].

Note

We chose 0 and 1. Any other pair of real numbers s and t will also work as long as – 4 < s < t < 4, f(s) > 0, and f(t) < 0.
The 0-and-1 pair is the simplest. 3.  Find the intervals in which the function f defined by f(x) = x2 + 4x + 3 is positive and negative.

Solution   4.  Show that the function f defined by f(x) = x2 – (a + b – 1)x + ab takes on the value (a + b)/2, where a and b are
any two real numbers.

Solution

We have f(x) = x2axbx + x + ab = (x a)(xb) + x.

Case a = b:  Replacing b with a we have f(x) = (xa)2 + x and (a + b)/2 = a. Thus, f(a) = (aa)2 + a = a =
(a + b)/2. If a > b, we'll just replace [a, b] by [b, a], and we'll reach the same conclusion.  Solution  