# Question Video: Intermediate Value Theorem

Intermediate Value Theorem Explained – To Find Zeros, Roots or C value – Calculus
Intermediate Value Theorem Explained – To Find Zeros, Roots or C value – Calculus

### Video Transcript

Let 𝑓 of 𝑥 equal three to the 𝑥
power minus 𝑥 to the fifth power. According to the intermediate value
theorem, which of the following intervals must contain a solution to 𝑓 of 𝑥 equals
zero? Is it the closed interval from two
to three, the closed interval from zero to one, the closed interval from negative
three to negative two, the closed interval from one to two, or the closed interval
from negative two to negative one?

So we have a function. How can we use the intermediate
value theorem to tell which of these intervals has a root of this function, a
solution to 𝑓 of 𝑥 equals zero? Well, let’s remind ourselves of the
intermediate value theorem. It states that if a function 𝑓 is
continuous on the closed interval from 𝑎 to 𝑏 and 𝑁 is some number between the
values of the function at the end points of that interval. That’s 𝑓 of 𝑎 and 𝑓 of 𝑏. Then there exists some number 𝑐 in
the open interval from 𝑎 to 𝑏, such that 𝑓 of 𝑐 equals 𝑁. The first thing to notice that our
function 𝑓 is continuous on the real numbers and so will be continuous on any of
the intervals in the options as well. So this hypothesis holds.

Now remember, we want to find a
solution to 𝑓 of 𝑥 equals zero. Comparing this with 𝑓 of 𝑐 equals
𝑁, it looks like we want to set 𝑁 equal to zero. So the intermediate value theorem
is telling us that for our continuous function 𝑓, if zero is between 𝑓 of 𝑎 and
𝑓 of 𝑏, then there exists 𝑐 in the open interval from 𝑎 to 𝑏 such that 𝑓 of 𝑐
is zero. In other words, if 𝑓 of 𝑎 and 𝑓
of 𝑏 have different signs, then there is some number 𝑐 between 𝑎 and 𝑏 which is
a root of 𝑓.

So to solve this question, we take
each interval in the options in turn, starting with the interval from two to
three. And if the signs of 𝑓 of two and
𝑓 of three are different, if one of them is positive and one of them is negative,
then we know there must be a root, a solution to 𝑓 of 𝑥 equal zero in this
interval. So let’s compute 𝑓 of two and 𝑓
of three. We do this using the definition for
𝑓 of 𝑥 we have in the question. We can evaluate these by hand or
using a calculator, finding that 𝑓 of three is negative 216 and 𝑓 of two is
negative 23. There’s no sign change of the
function here. Both values are negative.

By the intermediate value theorem,
we know that the function 𝑓 must take all values between negative the 23 and
negative 216 as its input changes from two to three. So we’d have a solution to 𝑓 of 𝑥
equals negative 100. For example, in this interval. However, as zero is not between
negative 23 and negative 216, we can’t say that there must be a solution to 𝑓 of 𝑥
equals zero in this interval.

We move on to option B. The closed interval from zero to
one. We compute the values of the
function at the end points. We find that 𝑓 of one is two and
𝑓 of zero is one. Again, there’s no sign change, so
no guarantee of a zero in this interval. However, we can see a sign change
between 𝑓 of one and 𝑓 of two. 𝑓 of one is positive and 𝑓 of two
is negative. The intermediate value theorem
tells us that as 𝑓 is continuous on the closed interval from one to two and as zero
is between 𝑓 of one, which is two, and 𝑓 of two, which is negative 23. Then there exists a number 𝑐 in
the open interval from one to two such that 𝑓 of 𝑐 is zero. And as 𝑐 lies in the open interval
from one to two, it must also lie in the closed interval from one to two. And so we have a solution to 𝑓 of
𝑥 equals zero in the closed interval from one to two. This is option D.

We can — if we like — check the
values of the function at the end point of the other intervals in the options to see
that there is no sign change in either intervals C or E. And so, D is definitely the only
correct answer. While the intermediate value
theorem guarantees us a root or solution to 𝑓 of 𝑥 equals zero in the interval
from one to two, we can’t say just based off the intermediate value theorem that
there are no roots in the other intervals.

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