Working with the Intermediate Value Theorem

At 6 a.m. the air temperature is 4 degrees Celsius, and at 2 p.m. it is 17 degrees. You can be sure that at some moment in between it was exactly 10 degrees, even with no thermometer reading at that moment, because temperature cannot get from 4 to 17 without passing through every value in between. The Intermediate Value Theorem (IVT) makes that reasoning precise for any continuous function. It is an existence theorem: it proves that a value is reached without telling you where.

What the theorem says

The picture explains it. Continuity on $[a, b]$ means the graph from $(a, f(a))$ to $(b, f(b))$ is a single unbroken curve. If the horizontal line $y = d$ lies between the heights of the two endpoints, one endpoint is below the line and the other above it, and an unbroken curve cannot get from one side to the other without meeting the line. Each place it meets the line is an input $c$ with $f(c) = d$.

This curve meets $y = d$ three times, at $c_1$, $c_2$ and $c_3$, while the theorem promises only one; the section after next says what it leaves out. A full proof relies on a property of the real numbers (that they have no gaps) studied in college analysis, and the AP exam treats the IVT as a theorem you apply, not one you prove.

The most-used special case is $d = 0$. If $f$ is continuous on $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs, then 0 is between them, so $f$ has at least one zero in $(a, b)$.

Continuity on the closed interval is the condition that matters

Every conclusion of the IVT depends on continuity on all of $[a, b]$, including the endpoints. Without it, the graph can skip over the line $y = d$.

• A jump skips values. Let $f(x) = x + 1$ for $x < 2$ and $f(x) = x + 3$ for $x \geq 2$. On $[0, 4]$, $f(0) = 1$ and $f(4) = 7$, and 4 is between them. But the left piece only produces values from 1 up to (not including) 3, and the right piece produces values from 5 to 7. No input gives 4. The IVT does not apply, because $f$ has a jump at $x = 2$.
• An asymptote skips values. On $\left[\frac{\pi}{4}, \frac{3\pi}{4}\right]$, $\tan\frac{\pi}{4} = 1$ and $\tan\frac{3\pi}{4} = -1$, opposite signs. Yet $\tan x$ has no zero on that interval ($\tan x = 0$ only where $\sin x = 0$, at multiples of $\pi$). The graph gets from positive to negative by going up to $\infty$ and coming back from $-\infty$ at the vertical asymptote $x = \frac{\pi}{2}$.

On the exam you confirm continuity in one of three ways:

• the problem says so ("$f$ is continuous on $[0, 12]$");
• the function is built from continuous pieces: polynomials, exponentials, $\sin x$ and $\cos x$ everywhere; rational functions, logarithms, roots and $\tan x$ on intervals inside their domains; and sums, products, quotients (with nonzero denominator) and compositions of these (topic 1.12);
• the problem says $f$ is differentiable, since a function that is differentiable at a point is continuous there (topic 2.4).

What the theorem does not say

The IVT is a one-way statement, and three tempting extensions of it are false.

It doesn't count. It guarantees at least one $c$, never exactly one. Claiming "the solution" or "exactly one zero" needs more information, such as knowing $f$ is always increasing (Unit 5).

It doesn't locate. It gives an interval that contains a $c$, not the value of $c$. You can shrink the interval by testing more inputs (Example 1 does this), but the IVT alone never names $c$.

It is silent when the value isn't bracketed. If $d$ is not between $f(a)$ and $f(b)$, the theorem makes no claim at all; in particular, if $f(a)$ and $f(b)$ have the same sign, it says nothing about zeros. The function might still have some. For $f(x) = 4x^2 - 1$ on $[-1, 1]$, both endpoint values are 3, and still $f\left(\pm\frac{1}{2}\right) = 0$. "The IVT does not guarantee a zero" and "there is no zero" are different statements.

Equations, tables, and the written justification

Equations between two functions. To show two continuous functions are equal somewhere, apply the IVT to their difference $h(x) = f(x) - g(x)$, which is continuous because $f$ and $g$ are. A sign change in $h$ between $a$ and $b$ means $f(c) = g(c)$ for some $c$ between them.

Tables. When a continuous function is given only by a table, apply the IVT on each pair of consecutive inputs. Each subinterval where $f$ passes from below $d$ to above it (or the reverse) contains its own solution of $f(x) = d$, and subintervals that don't overlap give different solutions. Counting those subintervals gives the fewest solutions the table guarantees.

Worked examples

Where students lose points

On the exam

The exam asks about the IVT in forms like Example 3: a table of values of a function stated to be continuous or differentiable, with a question such as "must there be a value..." or "what is the fewest number of times...." The answer earns credit only with the three-part justification: continuity, the inequality bracketing the value with numbers from the table, and the theorem's name with the conclusion. The work is the same with or without a calculator.