Intermediate Value Theorem

The Intermediate Value Theorem also called IVT, is a theorem in calculus about values that continuous functions attain between a defined interval. It guarantees the existence of a point within a continuous function's interval where the function takes on a specific value.

Statement of the Intermediate Value Theorem

The Intermediate Value Theorem states that,

For a continuous function f defined on the closed interval [a,b], where f(a) ≠ f(b) and d is real number between f(a) and f(b), then there exists at least one real number c between in the interval (a,b) such that f(c) = d.

To understand this better, imagine drawing the graph of f(x) on paper without lifting your pencil. If the graph starts at f(a) and ends at f(b), then at some point, it must pass through every value between f(a) and f(b).

Special Case of IVT: When f(a) and f(b) have opposite signs, then there exists at least one real number c between a and b such that f(c) = 0.

Converse of the Intermediate Value Theorem

The converse of the Intermediate Value Theorem (IVT) is not always true. The converse statement is stated as follows:

If there exists a point c ∈ [a, b] such that f(c) = d for every number d between f(a) and f(b), then f is continuous for the interval [a, b].

The above statement is not true always. A function can follow the Intermediate Value Theorem despite being discontinuous. In other words, a function following the IVT property need not to be a continuous function but a continuous function always follows the Intermediate Value Theorem.

Intermediate Value Theorem Proof

The theorem can be proved by two common approaches. Both are discussed one by one as follows.

Proof of IVT Based on Real Numbers' Completeness

The completeness of real numbers is the property which makes real numbers distinguishable from rational numbers and natural numbers. It implies that in a given interval of real numbers, each number is a real number.

The supremum or least upper bound states that for any set S which is subset of R which has an upper bound, there exists a smallest real number sup S that is greater than or equal to every element in S.

Now, we have a continuous function f defined on the interval [a, b], and d lies between f(a) and f(b). To prove that that there exists c ∈ [a, b] such that f(c) = d, we define a set as follows,

S = {x ∈ [a, b] | f(x) ≤ d}

Now, the upper bound of set S is b, as x ≤ b for all x∈[a, b]. By the completeness property of the real numbers, S has a supremum, say c = sup S, which implies that c ∈ [a,b]. We need to prove that f(c) = d.

Let f(c) ≠ d, then two cases arise,

• If f(c) < d, it implies that there exists an ε>0 such that for all x∈(c, c+ε)∩[a, b], f(x)<d. This contradicts c = sup S as it implies there are points greater than c but still in S.
• If f(c) > d, it implies that there exists an ε>0 such that for all x∈(c-ε, c)∩[a, b], f(x)<d. This contradicts c = sup S as it implies f takes values less than d for points close to c from the left.

From the contradictions proved above, it comes out that f(c) = d. Hence, we have proved that there exists c ∈ [a, b] such that f(c) = d, which proves the Intermediate Value Theorem.

Proof of IVT Using Continuity and Supremum

The theorem can also be proved using the property of continuity and supremum. The proof is similar to the previous one focusing on properties of continuous functions and supremum of a set.

Let us define a set S as follows,

S = {x ∈ [a, b] | f(x) ≤ d}

Let the supremum of set S is given as follows,

c = sup S

Where, c ∈ [a, b]

Now, let f(c) ≠ d, then two cases arise,

• f(c) < d, it implies that there exists a ε>0 such that if ∣x - c∣<ε, then ∣f(x)−f(c)∣<d−f(c). But this would imply that f(x)<d for x ∈ (c ,c+ε) ∩ [a, b], which contradicts c = sup S as there would be points greater than c with f(x) ≤ d.
• f(c) > d, it implies that there exists a ε>0 such that if ∣x - c∣<ε, then ∣f(x)−f(c)∣<f(c)-d, but this would imply that f(x)>d for x ∈ (c−ε, c) ∩ [a, b]. Here, x < c and c = sup S, which means f(x) ≤ d that contradicts the possibility of f(x) > d.

We get f(c) = d from the contradictions proved above, which proves the Intermediate Value Theorem (IVT).

Intermediate Value Theorem Applications

Various applications of the Intermediate Value Theorem are discussed as follows:

• The theorem can be used to check the existence of roots of a continuous function in a specified interval.
• The theorem helps to check whether a continuous function attains a given value in a defined interval.
• The theorem is used in solving equations through numerical methods such as Bisection Method.
• The theorem serves as a base for proving several other theorems in calculus such as Mean Value Theorem.
• The theorem is also used to find critical points, i.e. points where the derivative of the function attains a specific value.

Limitations of Intermediate Value Theorem

Although, Intermediate Value Theorem has various applications but it has some limitations too. These are discussed as follows:

• The theorem does not provide any information on number of roots in the interval for the function.
• The theorem is applicable to continuous functions only and is uncertain for discontinuous or piecewise functions.
• The theorem does not give any information about the behaviour of function outside the specified interval.

Also read

• Mean Value Theorem
• Cauchy's Mean Value Theorem
• Continuity of Functions

Solved Examples: Intermediate Value Theorem

Example 1: Check whether the function defined as f(x) = x³ - 8 has a root in the interval [0,4].

Solution:

Here, we have,

• f(0) = 0 - 8 = -8, and
• f(4) = 4³ - 8 = 64 - 8 = 56.

As the function yields values with opposite signs at the endpoints of the given interval, by intermediate value theorem, it implies that the function has at least one root in the interval.

Example 2: Show that the function defined as f(x) = e^x - 3x has a root in the interval [0,1].

Solution:

To show whether the function has a root in the given interval, we check the value of function at the endpoints of the interval,

We have, f(0) = e⁰ - 3(0) = 1 - 0 = 1, and

f(1) = e - 3 = -0.28 (approx.)

Thus, we see that function has opposite signed values at endpoints of the interval, so it has at least one root in the interval.

Practice Questions: Intermediate Value Theorem

Q1: Check whether the function defined as f(x) = x² - 2x has a root in the interval [0, 1].

Q2: Show that the function defined by f(x) = 1 - 2sin(x) has at least one root in the interval [0, π/2].

Q3: Consider the function f(x) = x³ - x + 2, check whether it has a root in the interval [1, 4].

Q4: Check whether the function given by f(x) = 4x - e^x has a root in the interval [0, 1].

Q5: Show that the function defined as f(x) = x⁵ - x has at least one root in the interval [-1, 1].