The Intermediate Value Theorem — AP® Calculus AB Notes

This is the first theorem of the course, and the first time the exam will ask you to justify an answer by naming a theorem and checking its conditions. The idea is simple. Getting the words right is what earns the points.

The idea

Suppose the temperature outside was $10^\circ$ C at 6 a.m. and $24^\circ$ C at noon. At some moment in between it was exactly $17^\circ$ C. Temperature does not jump; it passes through every value on its way. A function whose graph has no breaks behaves the same way.

Left: a continuous function on the closed interval [1, 5] with f(1) = 1 and f(5) = 5. The dashed line y = 3 lies between 1 and 5, and the curve crosses it three times, at c₁, c₂ and c₃, all inside (1, 5): the theorem promises at least one. Right: a function with a jump at x = 3 that also has f(1) = 1 and f(5) = 5 but skips over y = 3, so no c gives f(c) = 3. — **Figure 1.** Left: continuous on $[1, 5]$ , so $y = 3$ is reached (three times here). Right: a jump skips $y = 3$ . LevlPrep original diagram.

The theorem

Intermediate Value Theorem. If $f$ is continuous on the closed interval $[a, b]$ , and $d$ is any number between $f(a)$ and $f(b)$ , then there is at least one number $c$ in $(a, b)$ with $f(c) = d$ .

The two conditions before "then" are the theorem's hypotheses. They are not decoration. If either one fails, the theorem says nothing at all.

Continuous on the closed interval $[a, b]$ . One jump, hole or asymptote anywhere in $[a, b]$ is enough to break it. The graph on the right of the figure has $f(1) = 1$ and $f(5) = 5$ but jumps straight over 3.
$d$ between $f(a)$ and $f(b)$ . Values outside that range might be reached (a graph can rise above both ends) but nothing guarantees it.

Read the conclusion carefully too. It says at least one $c$ , and only that $c$ is somewhere in $(a, b)$ . The left graph in the figure crosses $y = 3$ three times. The theorem predicted one; it does not count them and does not locate them.

Writing the justification

Scoring guidelines give the point only when the continuity hypothesis is stated. A complete justification has four parts: why $f$ is continuous on the closed interval, the number $d$ between the two endpoint values, the theorem's name, and the conclusion.

Worked example 1. Show that $f(x) = x^3 - 4x + 1$ has a zero between 0 and 1.

Continuity: $f$ is a polynomial, so it is continuous on $[0, 1]$ .

Between: $f(0) = 1$ and $f(1) = 1 - 4 + 1 = -2$ , so $f(1) < 0 < f(0)$ .

Conclusion: by the Intermediate Value Theorem, $f(c) = 0$ for some $c$ in $(0, 1)$ .

Compare two weaker answers. " $f(0) > 0$ and $f(1) < 0$ , so $f$ has a zero between them" never mentions continuity: no credit for the reasoning. " $f$ is continuous, so it equals 0 somewhere" never shows 0 is between the endpoint values: also incomplete.

Using a table

Many exam questions give a continuous function only through a table. Apply the theorem to each pair of neighboring inputs.

Worked example 2. $T$ is continuous, with $T(0) = 20$ , $T(2) = 35$ , $T(5) = 31$ , $T(7) = 44$ , $T(10) = 40$ . What is the least number of times in $(0, 10)$ at which $T(t) = 33$ ?

On $[0, 2]$ : $20 < 33 < 35$ , so the IVT gives a time in $(0, 2)$ .

On $[2, 5]$ : $31 < 33 < 35$ , so another time in $(2, 5)$ .

On $[5, 7]$ : $31 < 33 < 44$ , so another in $(5, 7)$ .

On $[7, 10]$ : 33 is not between 44 and 40, so no guarantee.

The intervals $(0, 2)$ , $(2, 5)$ and $(5, 7)$ do not overlap, so $T(t) = 33$ at least 3 times.

Existence, then location

The theorem tells you a solution exists. To find it, use algebra if you can, or a graphing calculator. For $x^3 + x - 5 = 0$ on $(1, 2)$ : $f(1) = -3$ and $f(2) = 5$ , and $f$ is a polynomial, so a zero exists by the IVT. A calculator's zero finder gives $x \approx 1.516$ , written to three decimal places. Use radian mode whenever trigonometry is involved: $\cos x = x$ has its solution at $x \approx 0.739$ in radians.

This pattern, state the hypotheses and then apply the theorem, comes back with every theorem in the course. Learn it here, where the theorem is easy.