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In calculus, the Intermediate Value Theorem (IVT) is a foundational existence theorem. Intuitively, it states that if a continuous curve connects two points on a graph, it must pass through every intermediate y-value between those points without skipping any level.

For A-Level Mathematics and calculus courses, the IVT forms the mathematical basis for root-location and numerical methods (such as the change-of-sign method, interval bisection, and decimal search). Exam questions frequently require students to formally prove the existence of solutions to equations within given intervals, identify where functions achieve specific values, and justify why continuity is essential.

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Core Theory

The Intermediate Value Theorem Definition

Let f(x) be a function that is continuous on the closed interval [a, b]. If k is any real number strictly between f(a) and f(b) (such that f(a) < k < f(b) or f(b) < k < f(a)), then there exists at least one number c in the open interval (a, b) such that f(c) = k.

The Two Essential Hypotheses

For the Intermediate Value Theorem to guarantee a solution, two conditions must hold:

Closed Interval: The interval under consideration must include its endpoints: [a, b].
Continuity: The function f(x) must be continuous at every point in [a, b] (no jump discontinuities, removable holes, or vertical asymptotes).

Special Case: Bolzano's Theorem & Root Finding (Location of Roots)

The most common application of the IVT in A-Level exams is proving the existence of a root (where f(c) = 0).

If f(x) is continuous on [a, b], and f(a) and f(b) have opposite signs (one is positive and one is negative, meaning $f(a) \cdot f(b) < 0$ ):

Zero (0) lies strictly between f(a) and f(b).
By the IVT (setting k = 0), there must exist at least one value $c \in (a, b)$ such that:

f(c) = 0

Exam Focus

Examiner Tip #1: Always State "Continuous Function" Explicitly

In exam questions asking you to "show that a root exists between x = a and x = b," evaluating f(a) and f(b) to show a sign change is not enough. You must write:

"Since f(x) is continuous on [a, b] and there is a change of sign, by the Intermediate Value Theorem, there exists at least one root $c \in (a, b)$ ."

Omitting the word continuous loses the justification mark across all UK exam boards.

Examiner Tip #2: Transforming Equations to f(x) = 0

When given an equation where terms appear on both sides (e.g., $e^x = 4 - x^2$ ), always rearrange all terms to one side first:

f(x) = e^x + x^2 - 4 = 0

Then evaluate f(a) and f(b) for sign change.

Examiner Tip #3: Asymptotes and False Roots

A change of sign does not guarantee a root if the function is discontinuous.

Example: For $f(x) = \dfrac{1}{x}$ , $f(-1) = -1$ and $f(1) = 1$ . Although there is a sign change across [-1, 1], f(x) = 0 has no real solution because f(x) has an infinite discontinuity at x = 0.

Worked Example 1: Proving the Existence of a Polynomial Root

Show that the equation $x^3 - 3x - 1 = 0$ has a real root in the interval $[1, 2]$ .

Step 1: Define the function and state continuity.

Let $f(x) = x^3 - 3x - 1$ .

$f(x)$ is a polynomial, and all polynomials are continuous for all $x \in \mathbb{R}$ , hence continuous on the closed interval $[1, 2]$ .

Step 2: Evaluate the function at the interval endpoints.

$f(1) = (1)^3 - 3(1) - 1 = 1 - 3 - 1 = -3$

$f(2) = (2)^3 - 3(2) - 1 = 8 - 6 - 1 = +1$

Step 3: Conclude using IVT.

Since $f(1) = -3 < 0$ and $f(2) = 1 > 0$ , there is a change of sign ( $f(1) < 0 < f(2)$ ).

Because $f(x)$ is continuous on $[1, 2]$ , by the Intermediate Value Theorem, there exists at least one real value $c \in (1, 2)$ such that $f(c) = 0$ .

Practice Questions & Solutions

Score:

Prove that the function

$\text{[math]}$

takes the value 6 at least once in the interval [1, 4].

Solution

Step 1: Establish continuity

$\text{[math]}$

is the sum of two continuous functions on [1, ∞), so g(x) is continuous on the closed interval [1, 4].

Step 2: Evaluate the function at the endpoints.

$\text{[math]}$

Step 3: Apply the IVT.

The target value k = 6 lies strictly between g(1) and g(4):

Since g(x) is continuous on [1, 4], by the Intermediate Value Theorem, there exists at least one number $\text{[math]}$ such that g(c) = 6.

For the function

$\text{[math]}$

on the interval [0, 3], verify that the IVT applies for k = 1, and find all values of $\text{[math]}$ such that f(c) = 1

Solution

Step 1: Check IVT conditions.

f(x) is a polynomial and therefore continuous on [0, 3].

$\text{[math]}$

Since f(0) = 1 lies within (0,3) the theorem applies.

Step 2: Solve f(c) = 1 algebraically.

$\text{[math]}$

Step 3: Select values within the open interval (0, 3).

While c = 0 is an endpoint, c = 2 lies strictly inside the open interval (0, 3).

Thus, the value guaranteed inside the interval is c = 2.

Show that the polynomial equation

$\text{[math]}$

has at least one real root in the closed interval [1, 2].

Solution

Step 1: Let

$\text{[math]}$ .

Since f(x) is a polynomial, it is continuous on all of $\text{[math]}$ , including the closed interval [1, 2].

Step 2: Evaluate f(x) at the interval endpoints:

$\text{[math]}$

Step 3: Since

f(1) = -5 <0

and

f(2) = 12 > 0

there is a sign change.

Step 4: By the Intermediate Value Theorem, because f(x) is continuous on [1, 2], there exists at least one value $\text{[math]}$ such that f(c) = 0.

For the function

$\text{[math]}$

on the interval [1, 3], verify that the Intermediate Value Theorem applies for the target value k = 3, and determine the exact value of $\text{[math]}$ where f(c) = 3.

Solution

Step 1: f(x) is a polynomial, hence continuous on [1, 3].

Step 2: Evaluate endpoints:

$\text{[math]}$

Step 3: Since f(1) <3 < f(3), k=3 lies between f(1) and f(3).< p>

By IVT, there exists $\text{[math]}$ such that f(c) = 3.

Step 4: Solve f(c) = 3:

$\text{[math]}$

Testing integer factors of -6: For c = 2:

$\text{[math]}$

For c = 3:

$\text{[math]}$

Factor by grouping/division: Testing c = 2 fails, but let's test other values or factor:

$\text{[math]}$

Notice

$\text{[math]}$

Using synthetic division with root near $\text{[math]}$ , or by factoring:

If rewritten as

$\text{[math]}$

the exact real root of

$\text{[math]}$

$\text{[math]}$ , which lies strictly in the interval (1, 3).

A rational function is given by

$\text{[math]}$

Explain why the change of sign between g(2) = -5 and g(4) = 9 does not guarantee that g(x) = 0 has a solution in the interval [2, 4].

Solution

Step 1: Evaluate the denominator of

$\text{[math]}$

Setting the denominator to zero yields $\text{[math]}$

Step 2:

The point x = 3 lies within the closed interval [2, 4]. At x = 3, g(x) is undefined and has a vertical asymptote (infinite discontinuity).

Step 3:

Because g(x) is not continuous throughout the entire interval [2, 4], the foundational hypothesis of the Intermediate Value Theorem is violated.

Step 4:

Consequently, a change of sign between g(2) = -5 and g(4) = 9 does not guarantee a root and solving

$\text{[math]}$

which is not in [2, 4]

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Gianpiero Placidi

UK-based Chemistry graduate with a passion for education, providing clear explanations and thoughtful guidance to inspire student success.

Formulas

Continuity Formulas

Limit Formulas

what will be sign of -infinity is to power infinity plus or minus

Vanessa - Editorial Manager Superprof UK

July 2026

The sign **cannot be determined**.

For a negative base:

* **Even exponent** → **positive**
* **Odd exponent** → **negative**

So as the exponent tends to infinity, the sign depends on whether the exponent is even or odd. Therefore, ((-\infty)^\infty) has **no definite sign**—it can be **positive or negative**.

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Chapters

Core Theory
Exam Focus
Worked Example 1: Proving the Existence of a Polynomial Root
Practice Questions & Solutions

Finding Intermediate Values: Practice Questions and Solutions

Core Theory