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Mathematical Expression Editor
In this section we apply a theoretically important existence theorem called the
Intermediate Value Theorem.
1 Intermediate Value Theorem
In this section, we introduce an important theorem related to continuous functions,
the Intermediate Value Theorem. To build intuition, let us consider two real-life
scenarios: the behavior of ambient temperature and the balance in a bank account.
Suppose the temperature is at 8 a.m. and at noon. Because temperature changes
continuously, we can be certain that at some time between 8 a.m. and noon, the
temperature was exactly . This conclusion follows from the continuous nature of
temperature: it does not jump abruptly from one value to another but passes
smoothly through all intermediate values. Now, consider a different scenario involving
a bank account. Suppose the account balance is $65 at 8 a.m. and $75 at noon. Can
we similarly conclude that the account balance was exactly $70 at some time
between 8 a.m. and noon? Unlike the temperature example, we cannot make
this claim with certainty. Here’s why: It is possible that a single deposit of
$10 was made at 11 a.m., causing the balance to jump directly from $65
to $75 without ever being exactly $70. On the other hand, if the $10 were
added in smaller increments (e.g., two deposits of $5 each), the account could
indeed have had a balance of $70 at some point. This uncertainty arises
because bank account balances can change discontinuously due to individual
transactions, which may create jumps in value. The fundamental difference between
these two scenarios lies in the concept of continuity. Temperature changes
continuously, while bank account balances can exhibit discontinuities. Continuity
ensures that if a quantity changes from one value to another, it must pass
through every intermediate value in between. In contrast, discontinuous
changes allow for abrupt jumps, skipping over intermediate values entirely.
Mathematically, this property is captured by the Intermediate Value Theorem
(IVT):
Intermediate Value Theorem If the function is continuous on the closed interval and
is a number strictly between and , then there exists a number in the open interval
such that . This means that the equation has at least one solution in the open
interval .
The IVT formalizes our intuition about continuous change. It guarantees that for
any continuous function, every intermediate value between and must occur at least
once within the interval . In the examples above, the temperature behaves like
a continuous function, making the IVT applicable and ensuring that the
value was reached. In contrast, the bank account balance behaves like a
discontinuous function, so the IVT cannot be applied, and we cannot draw the same
conclusion. The value in the theorem is called an intermediate value for the function on the
interval .
The following figure illustrates the IVT.
example 1 Show that the equation has a solution between and . First, the function is continuous on the interval since is a polynomial. Second,
observe that and so that 10 is an intermediate value, i.e., Now we can apply the
Intermediate Value Theorem to conclude that the equation has a least one solution
between and . In this example, the number 10 is playing the role of in the statement
of the theorem.
(problem 1) Determine whether the IVT can be used to show that the equation has a
solution in the open interval .
Is continuous on the closed interval ?
YesNo
and
Is an intermediate value?
YesNo
Can we apply the IVT to conclude that the equation has a solution in the open
interval ?
YesNo
example 2 Show that the equation has a solution between and .
First, note that the function is continuous on the interval and hence it is
continuous on the sub-interval, . Next, observe that and so that 2 is an
intermediate value, i.e., Finally, by the Intermediate Value Theorem we
can conclude that the equation has a solution on the open interval . In
this example, the number 2 is playing the role of in the statement of the
theorem.
(problem 2) Determine whether the IVT can be used to show that the equation has a
solution in the open interval Is continuous on the closed interval ?
YesNo
and
Is an intermediate value?
YesNo
Does the IVT imply that the equation has a solution in the open interval ?
YesNo
example 3 Show that the function has a root in the open interval .
Recall that a root of a polynomial occurs when . Since is a polynomial, it is
continuous on the interval . Plugging in the endpoints shows that 0 is an intermediate
value: and so By the IVT, we can conclude that the equation has a solution (and
hence has a root) on the open interval .
(problem 3a) Determine whether the IVT can be used to show that the function has
a root in the open interval .
A root of a polynomial is a solution to the equation
Is continuous on the closed interval ?
YesNo
and
Is an intermediate value?
YesNo
Does the IVT imply that the function has a root in the open interval ?
YesNo
(problem 3b) Determine whether the IVT can be used to show that the function has
a zero in the open interval .
A zero of a function is a solution to the equation
Is continuous on the closed interval ?
YesNo
and
Is an intermediate value?
YesNo
Does the IVT imply that the function has a zero in the open interval ?
YesNo
example 4 Show that the equation has a solution between and . Note that the equation is equivalent to the equation . The latter is the preferred
form for using the IVT. So let Since is the difference between two continuous
functions, it is continuous on the closed interval . Next, we compute and and show
that 0 is an intermediate value: and and so, By the IVT, the equation has a
solution in the open interval . Hence the equivalent equation has a solution on the
same interval.
(problem 4) Use the IVT to show that the equation has a solution in the open
interval .
Rewrite the equation in the form
The functions and are continuous for all
example 5 Use the IVT four times to approximate a root of the polynomial First, is
a polynomial, so it is continuous on any closed interval. Next, note that By the IVT
has a root in the interval . To use the IVT a second time, we now determine the
midpoint of this interval: and we plug this in to : Combining this with we can use
the IVT again to conclude that has a root on the interval . This interval
is half of the original interval- the original interval has been bisected. We
now use the IVT a third time. The midpoint of the interval is . We plug
this into : Combining this with we can use the IVT to conclude that has
a root on the interval . This interval is half of the previous interval- the
previous interval has been bisected. We will do this a fourth and final time.
Note that is the midpoint of the interval and Combining this with we
can use the IVT a fourth time to conclude that has a root on the interval
. At this stage, our approximation of the root is which is the midpoint
of this interval. Our error is then no more than , which is half the width
of the interval. We will stop here, but the method could theoretically be
continued indefinitely giving a better approximation to the root each time.
This method of approximating roots is called the Method of Continued
Bisection.
Here is a detailed, lecture style video on the Intermediate Value Theorem: