The Intermediate Value Theorem

What you will be able to do

• state the Intermediate Value Theorem and say why each condition is needed;
• show that an equation has a root in an interval by a change of sign;
• find a root by halving the interval, and count how many halvings are needed.

The theorem

Walk from a valley at height $f(a)$ up to a hill at height $f(b)$ along a path with no jumps. You must pass through every height in between. This is the Intermediate Value Theorem.

If $f$ is continuous on $[a, b]$ and $N$ is any number between $f(a)$ and $f(b)$, then there is at least one number $c$ in $(a, b)$ with $f(c) = N$.

The special case $N = 0$ is the one used most. If $f$ is continuous on $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs, then the equation $f(x) = 0$ has a root in $(a, b)$.

Why continuity is needed

Take $f(x) = \dfrac{1}{x}$ on $[-1, 1]$. Then $f(-1) = -1$ and $f(1) = 1$ have opposite signs. But $\dfrac{1}{x}$ is never zero. The theorem does not apply, because $f$ has an infinite break at $x = 0$ inside the interval.

What the theorem does not say

• No sign change does not mean no root. The function $x^2 - 3x + 2$ is positive at $x = 0$ and at $x = 3$, yet it is zero at $x = 1$ and $x = 2$.
• A sign change does not mean exactly one root. It guarantees at least one. There could be three.
• The theorem says a point exists. It does not say where it is.

Finding the root by bisection

The theorem gives a method. Suppose $f$ is continuous on $[a, b]$ with a sign change.

1. Find the midpoint $m = \dfrac{a + b}{2}$ and the sign of $f(m)$.
2. Keep the half of the interval where the sign still changes.
3. Repeat.

Each step halves the interval. After $n$ steps its length is $\dfrac{b - a}{2^n}$. To start with an interval of length $1$ and finish with one shorter than $0.01$, you need $2^n > 100$. That is $n = 7$.

A worked example

Show that $x^3 + x - 1 = 0$ has a root between $0$ and $1$, and trap it more tightly.

1. Continuity. $f(x) = x^3 + x - 1$ is a polynomial, so it is continuous everywhere.
2. A sign change. $f(0) = -1$ and $f(1) = 1$. By the Intermediate Value Theorem, a root lies in $(0, 1)$.
3. Halve. $f(0.5) = -0.375$, so the root is in $(0.5, 1)$. Then $f(0.75) = 0.171875$, so it is in $(0.5, 0.75)$.
4. Halve again. $f(0.625) \approx -0.1309$, so the root is in $(0.625, 0.75)$. Then $f(0.6875) \approx 0.0125$, so it is in $(0.625, 0.6875)$.

After four halvings the root is trapped in an interval of length $\dfrac{1}{16}$.

Practise with new numbers

Try it in Python

The program carries out the halving for $x^3 + x - 1 = 0$. After each step it prints the interval that still holds the root.

Common mistakes

• Using the theorem without continuity. Always say that the function is continuous on the closed interval.
• Concluding no root from equal signs. Two roots can sit between two positive values.
• Taking the sign change as one root. The theorem gives at least one.
• Checking only open ends. The theorem uses the values $f(a)$ and $f(b)$ at the ends of a closed interval.

Remember these

• If $f$ is continuous on $[a, b]$ and $N$ lies between $f(a)$ and $f(b)$, then $f(c) = N$ for some $c$ in $(a, b)$.
• A sign change of a continuous function traps a root.
• Bisection halves the interval each time, so $n$ steps leave a length of $\dfrac{b - a}{2^n}$.
• No sign change proves nothing.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A function $f$ is continuous on $[1, 4]$ with $f(1) = -2$ and $f(4) = 6$. Must $f$ take the value $3$? What about the value $7$?

2. Show that $x^2 - 5 = 0$ has a root in $(2, 3)$.

3. The function $\dfrac{1}{x}$ has opposite signs at $-1$ and $1$ but no root between. Why does this not break the theorem?

4. The function $x^2 - 3x + 2$ is positive at $0$ and at $3$. Does the equation $x^2 - 3x + 2 = 0$ have a root in $(0, 3)$?

5. Bisection starts with an interval of length $1$. How many halvings leave an interval shorter than $0.01$?

References

• OpenStax, Calculus Volume 1, Chapter 2, section on continuity (the Intermediate Value Theorem).