Video Transcript
If 𝑓 of 𝑥 is continuous over the
closed interval from zero to three, 𝑓 of zero is greater than zero, and 𝑓 of three
is greater than zero, can we use the intermediate value theorem to conclude that 𝑓
of 𝑥 has no zeros in the interval from zero to three?
Let’s have a go at sketching this
scenario on a graph. We know that 𝑓 of zero is
positive. Let’s draw it here and 𝑓 of three
is also positive. So maybe, the graph goes through
this point and 𝑓 of 𝑥 is continuous. Does this mean that 𝑓 of 𝑥 has no
zeros in the interval from zero to three? Well, no. We can sketch a graph of a
continuous function 𝑓 for which both 𝑓 of zero and 𝑓 of three are positive, but
which has zeros in the closed interval from zero to three. So we’d hope that we can’t use the
intermediate value theorem to conclude that 𝑓 of 𝑥 has no zeros because it simply
isn’t true.
Why might we think that the
intermediate value theorem implies this incorrect statement? What the intermediate value theorem
states is that if 𝑓 is continuous on the closed interval from 𝑎 to 𝑏 and 𝑁 is
the number between 𝑓 of 𝑎 and 𝑓 of 𝑏, then there exists a number 𝑐 in the open
interval from 𝑎 to 𝑏 such that 𝑓 of 𝑐 equals 𝑁. Setting 𝑁 equal to zero, we get
the special case that if 𝑓 is a continuous function and 𝑓 of 𝑎 and 𝑓 of 𝑏 have
opposite signs, then there exists a number 𝑐 in the open interval from 𝑎 to 𝑏
such that 𝑓 of 𝑐 equals zero. In other words, there is a zero of
the function 𝑓 in the interval.
This special case is sometimes
known as Bolzano’s theorem. We have to be careful here. The intermediate value theorem does
not mean that if 𝑁 is not between 𝑓 of 𝑎 and 𝑓 of 𝑏, then there does not exist
a number 𝑐 in the open interval from 𝑎 to 𝑏 such that 𝑓 of 𝑐 equals 𝑁. So the special case does not mean
that if 𝑓 of 𝑎 and 𝑓 of 𝑏 have the same sign. In other words, if they’re both
positive or both negative, then there cannot exist a number 𝑐, which is a root of
𝑓. This is a statement that we’re
asked about in the question and it does not follow from the intermediate value
theorem. Our answer is, therefore, no. We cannot conclude that 𝑓 of 𝑥
has no zeros in the interval from zero to three.
