The Intermediate Value Theorem

Below at one end, above at the other

Take f(x) = x² − 2. At x = 0 it is −2, below the x-axis. At x = 2 it is 2, above the axis. f is a polynomial, so it is continuous, and its graph from x = 0 to x = 2 is one unbroken curve.

y = x² − 2 from x = 0 to 2.5. It starts at (0, −2), below the axis, and passes (2, 2), above it. In between it crosses the axis once, at x = √2 = 1.414.

It must cross zero

A curve drawn without lifting the pencil cannot get from below the axis to above it without touching the axis on the way. Only a break, such as a jump, could let it skip from one side to the other.

The intermediate value theorem says this precisely. If f is continuous on the closed interval from a to b, and N is any number between f(a) and f(b), then there is at least one c between a and b with f(c) = N. A continuous function takes every value between its values at the two ends.

For x² − 2 on the interval from 0 to 2, f(0) = −2 and f(2) = 2, and 0 lies between them, so there is a c with c² − 2 = 0. It is c = √2. The same reasoning with N = 5 on the interval from 0 to 4, where f(4) = 14, gives a c with c² − 2 = 5, which is c = √7 = 2.646.

A sign change traps a root

With N = 0 the theorem becomes a test for roots: if f is continuous on the interval from a to b and f(a) and f(b) have opposite signs, then f has a root between a and b.

The theorem says that a root exists. It does not say where, but repeated sign changes narrow it down. For x² − 2: f(1) = −1 and f(1.5) = 0.25, so the root is between 1 and 1.5. f(1.4) = −0.04 and f(1.42) = 0.0164, so it is between 1.4 and 1.42. f(1.41) = −0.0119 and f(1.415) = 0.002225, so it is between 1.41 and 1.415, and the root is 1.41 to two decimal places.

Every numerical root-finder rests on this. Two values of opposite sign, from a continuous function, trap a root between them.

At least one, not exactly one

The theorem promises at least one c, and there may be more. f(x) = x³ − x on the interval from −2 to 2 has f(−2) = −6 and f(2) = 6, a sign change, and it has three roots in between: x³ − x = x(x − 1)(x + 1) is 0 at x = −1, 0 and 1.

To say there is exactly one root needs something more, such as the function only ever rising or only ever falling on the interval. x² − 2 only rises for x ≥ 0, which is why it crosses once.

y = x³ − x from x = −2 to 2. It goes from (−2, −6) to (2, 6) and crosses the axis three times, at x = −1, 0 and 1.

Continuity on the whole closed interval

Without continuity the theorem says nothing. f(x) = 1/x has f(−1) = −1 and f(1) = 1, a sign change, yet 1/x is never 0. It is not continuous on the interval from −1 to 1: at x = 0 it has no value, and the curve runs off along an asymptote instead of crossing the axis.

The continuity must hold at the ends of the interval as well. Let f(0) = −1 and f(x) = 1 for 0 < x ≤ 1. Then f(0) = −1 and f(1) = 1, and f is continuous at every point between 0 and 1, but not at 0 itself: the values just right of 0 are all 1, not −1. f is never 0.

y = 1/x from x = −1.5 to 1.5. It is −1 at x = −1 and 1 at x = 1, but it never meets the axis: it breaks at the dashed asymptote x = 0.

f(0) = −1, and f(x) = 1 for 0 < x ≤ 1. The solid dot at (0, −1) is the value at 0; the hollow dot at (0, 1) is where the rest of the graph begins without including it. The values go from −1 to 1 and never pass 0.

No sign change does not mean no root

The theorem only works in one direction. If f(1) = 2 and f(2) = 6, it forces every value between 2 and 6, and 0 is not one of them, so no root is forced. There may still be roots: f(x) = x² − 1 has f(−2) = 3 and f(2) = 3, no sign change, and roots at x = −1 and x = 1.

The usual mistakes

Using the theorem on a function that is not continuous on the whole closed interval. A step or an asymptote can skip from one sign to the other.

Thinking the curve must be a straight line. Any continuous curve from −3 up to 5 passes through 0 on the way, however it bends.

Claiming exactly one root from a sign change. The theorem gives at least one; x³ − x has three between −2 and 2.

Claiming no root because there is no sign change. x² − 1 has the same sign at −2 and 2 and two roots between them.

Using it for a value N outside the range from f(a) to f(b). Then the theorem says nothing either way.

An oven cooling and a balance that jumps

In the first application below, a cooling oven is shown to pass exactly 100 degrees at some moment: the theorem gives the moment, and the fact that the oven only cools makes it the only one. In the second, a bank balance goes from below $800 to above it without ever being $800, because it changes by a jump.

Worked example: A Bread Oven Cooling Down: The Minute It Passes a Hundred Degrees

Question A bread oven is switched off and cools. Its temperature m minutes later is T(m) = 20 + 200e^−m/25 degrees Celsius, for 0 ≤ m ≤ 60. (a) Use the intermediate value theorem to show that the oven is at exactly 100 degrees Celsius at some time in the hour, and explain why there is exactly one such time. (b) Find that time, in minutes to one decimal place.

1.T is built from an exponential and two constants, so it is continuous on the whole closed interval from m = 0 to m = 60. That is the first condition.

T is built from an exponential and two constants, so it is continuous on the whole closed interval from m = 0 to m = 60.
2.Work out the two end values: T(0) = 20 + 200 = 220 and T(60) = 20 + 200e^−2.4 = 38.14 to two decimal places.

The end values are T(0) = 220 and T(60) = 20 + 200e^−2.4 = 38.14 degrees Celsius.
3.(a) 100 lies between 38.14 and 220, so the intermediate value theorem gives a time m in the interval with T(m) = 100. The term 200e^−m/25 falls as m rises, so T is strictly decreasing and passes 100 once and no more.

(a) 100 lies between 38.14 and 220, so the intermediate value theorem gives a time with T(m) = 100; T is strictly decreasing, so there is exactly one.
4.For part (b), solve 20 + 200e^−m/25 = 100. Take 20 from both sides to get 200e^−m/25 = 80, then divide both sides by 200 to get e^−m/25 = 0.4.

For part (b), 20 + 200e^−m/25 = 100 gives 200e^−m/25 = 80, so e^−m/25 = 0.4.
5.Take logarithms of both sides: −m25 = ln 0.4, so m = 25 ln 2.5. (b) m = 22.9 minutes to one decimal place. Check: T(22.9) = 100.02 degrees Celsius, which is 100 to the nearest degree.

(b) Taking logarithms, m = 25 ln 2.5 = 22.9 minutes, and T(22.9) = 100.02 degrees Celsius.

Answer: (a) T is continuous on the interval, T(0) = 220 and T(60) = 38.14, and 100 lies between them, so the theorem gives a time with T(m) = 100; T is strictly decreasing, so there is exactly one such time; (b) m = 25 ln 2.5 = 22.9 minutes

Common mistakes

Using the theorem on an interval where the target does not lie between the end values. The theorem promises a crossing only when the target is between T(a) and T(b); outside that range it says nothing either way.
Claiming from the theorem alone that the time is unique. The theorem gives at least one crossing; uniqueness here comes from the extra fact that T is strictly decreasing.

More continuity problems, worked step by step →

Worked example: A Current Account That Skips Eight Hundred Dollars: A Theorem Whose Condition Fails

Question An account holds $420 from the first of the month. On the tenth a payment of $600 is credited and the balance is $1020 for the rest of the month. The balance is below $800 on the ninth and above $800 on the eleventh, and yet the account is never at exactly $800. (a) Explain this, naming the condition of the intermediate value theorem that fails. (b) Find the payment that would have left the balance at exactly $800 on the tenth.

1.Write the balance as a function of the day: B(d) = 420 for d < 10, and B(d) = 1020 for d ≥ 10. Those are the only two values it takes all month.

As a function of the day, B(d) = 420 for d < 10 and B(d) = 1020 for d ≥ 10. Those are the only two values it takes.
2.At the payment, lim_{d → 10⁻} B(d) = 420 and lim_{d → 10⁺} B(d) = 1020, so B has a jump discontinuity at d = 10 of size 1020 − 420 = 600.

At the payment, lim_{d → 10⁻} B(d) = 420 and lim_{d → 10⁺} B(d) = 1020: a jump discontinuity of size 600.
3.(a) The intermediate value theorem asks for a function continuous on the closed interval. B is not continuous at d = 10, so the theorem does not apply and its conclusion fails: the balance takes no value strictly between $420 and $1020, and $800 is one of those values.

(a) The intermediate value theorem asks for a continuous function. B is not continuous at d = 10, so no value strictly between $420 and $1020 is ever taken.
4.For part (b), the balance on the tenth is the old balance plus the payment, so a payment of p dollars leaves 420 + p dollars.

For part (b), a payment of p dollars leaves 420 + p dollars on the tenth.
5.(b) Set 420 + p = 800, so p = $380. Check: 420 + 380 = 800. The account reaches $800 by landing on it, never by crossing it.

(b) 420 + p = 800 gives p = $380. The balance reaches $800 by landing on it, never by crossing it.

Answer: (a) the balance is not continuous: it jumps from $420 to $1020 on the tenth, a jump of $600, and the theorem needs a continuous function, so every value between them is skipped; (b) $380

Common mistakes

Treating the theorem as a fact about any quantity that goes from below a value to above it. Without continuity there is no crossing to find, and a step record is the everyday example of that.
Reading the failure as the theorem being wrong. The theorem is stated for continuous functions, and here one of its conditions is simply not met, so it says nothing about this account.

More continuity problems, worked step by step →