AP Calculus AB · Study guide

Working with the Intermediate Value Theorem (IVT)

AP Calculus · Limits and Continuity

Working with the Intermediate Value Theorem

Three hypotheses, one conclusion, and a justification the reader expects word for word.

key idea$\text{continuous on }[a,b]\ \Rightarrow\ f\text{ takes every value between }f(a)\text{ and }f(b)$

The theorem

Intermediate Value Theorem

Hypotheses — all must hold

$f$ is continuous on the closed interval $[a,b]$.
$L$ is a number strictly between $f(a)$ and $f(b)$.

What you may then conclude

There exists at least one $c$ in the open interval $(a,b)$ with $f(c)=L$.

Drop continuity and the conclusion dies

The step function below has $f(0)=-1.4$ and $f(2)=1.8$, so $L=0$ lies between them — yet the function jumps straight over $0$ and never equals it. Continuity is not decoration; it is what forbids the graph from skipping a value.

[Figure: continuous curve crossing the target value]

$g(x)=x^{3}-2x-1$ is continuous on $[0,2]$ with $g(0)=-1$ and $g(2)=3$. Since $0$ lies between them, some $c$ in $(0,2)$ has $g(c)=0$.

Checking the hypotheses

Nearly all the marks on an IVT question are for the justification, not the answer. Say the three things below explicitly, in this order, every time.

The three-sentence justification

State continuity and why. “$f$ is a polynomial, so it is continuous on $[a,b]$.” Rational, root and trig functions need a domain check first.
Evaluate both endpoints. Give the actual numbers: $f(a)=\dots$ and $f(b)=\dots$.
Locate $L$ between them and invoke the theorem. “Since $f(a)\lt L\lt f(b)$, by the Intermediate Value Theorem there exists $c$ in $(a,b)$ with $f(c)=L$.”

[Figure: step function that skips the target value]

The same endpoints, but with a jump at $x=1$. Every hypothesis except continuity still holds, and the conclusion fails outright.

Worked examples

EX 1

Show a root exists

$x^{3}-2x-1$ on $[0,2]$

Show that $g(x)=x^{3}-2x-1$ has a root in $(0,2)$.

Continuity

$g$ is a polynomial, so it is continuous on the closed interval $[0,2]$.

Endpoints

$g(0)=-1$ and $g(2)=8-4-1=3$.

Locate $L$

We want a root, so $L=0$. Since $-1\lt 0\lt 3$, the value $0$ lies strictly between $g(0)$ and $g(2)$.

Invoke the theorem

By the IVT there exists $c$ in $(0,2)$ with $g(c)=0$. Hence $g$ has a root there.

What you have NOT shown

The IVT gives existence only. It does not say the root is unique, and it does not locate it.

EX 2

Check the domain before claiming continuity

trap

A student argues that $f(x)=\dfrac{1}{x-1}$ satisfies the IVT on $[0,2]$ because $f(0)=-1$ and $f(2)=1$, so $f(c)=0$ for some $c$. What is wrong?

Test the hypothesis, not the endpoints

The endpoint values are computed correctly, but the hypothesis is about the whole interval.

Find the break

$f$ is undefined at $x=1$, which lies inside $[0,2]$. So $f$ is not continuous on $[0,2]$ and the IVT does not apply.

Check the claim independently

$\tfrac{1}{x-1}=0$ has no solution — a reciprocal is never zero. The conclusion is not merely unjustified, it is false.

Lesson

For rational functions, always confirm the denominator has no zero inside the interval before invoking the IVT.

EX 3

Reading it from a table

table

A continuous function $h$ has $h(1)=4$, $h(3)=-2$, $h(6)=5$. What is the minimum number of zeros $h$ must have on $[1,6]$?

First subinterval

On $[1,3]$, $h$ goes from $4$ to $-2$. Since $-2\lt 0\lt 4$, the IVT guarantees a zero in $(1,3)$.

Second subinterval

On $[3,6]$, $h$ goes from $-2$ to $5$. Again $0$ lies between, so there is a zero in $(3,6)$.

Count

The two subintervals are disjoint, so these are two different zeros. $h$ must have at least two zeros on $[1,6]$.

Why ‘at least’

$h$ could wiggle and cross many more times. The IVT sets a floor on the count, never a ceiling.

EX 4

Use it on a difference of functions

$f(x)=g(x)$

Show that $\cos x = x$ for some $x$ in $\left(0,\tfrac{\pi}{2}\right)$.

Turn the equation into a root problem

Let $d(x)=\cos x-x$. A solution of $\cos x=x$ is exactly a zero of $d$.

Continuity

$\cos x$ and $x$ are both continuous everywhere, and a difference of continuous functions is continuous. So $d$ is continuous on $\left[0,\tfrac{\pi}{2}\right]$.

Endpoints

$d(0)=1-0=1\gt 0$ and $d\!\left(\tfrac{\pi}{2}\right)=0-\tfrac{\pi}{2}\approx-1.571\lt 0$.

Invoke

Since $d$ changes sign, by the IVT there is $c$ in $\left(0,\tfrac{\pi}{2}\right)$ with $d(c)=0$, i.e. $\cos c=c$.

Pattern

“Show $f=g$ somewhere” almost always means “apply the IVT to $f-g$.”

When it fails

The interval must be closed and continuity must hold on all of it

A single hole, jump or asymptote anywhere inside $[a,b]$ voids the theorem — even if the endpoints look fine, as in Example 2.

$L$ strictly between, and $c$ strictly inside

The theorem places $c$ in the open interval $(a,b)$. Writing “$c$ in $[a,b]$” is a weaker, sloppier claim and can cost the justification mark.

Existence only — never uniqueness

The IVT never says ‘exactly one’. If a question asks for uniqueness you need a monotonicity argument (e.g. $f'\gt 0$) on top.

The converse is not true

A function can take every intermediate value and still be discontinuous. Hitting all the values does not prove continuity.

Course map. Topic 1.16 · LO FUN-1.A — explain the behaviour of a function on an interval using the Intermediate Value Theorem (EK FUN-1.A.1).