hi everyone today we're going to talk about how to use the intermediate value theorem
to prove that the function has a root in the given interval to complete problem we'll evaluate
the function at both ends of the interval and then show that the graph of the function
crosses the x axis between these values let's take a look in this particular problem we've
been asked to use the intermediate value theorem to show that the function f of x equals x
to the fourth plus x minus three has a root on the interval one to two so the first thing
we want to make clear is that the root of a function is a point at which the graph of
the function crosses the x axis in order to use the intermediate value theorem to show
that this function crosses the x axis between one and two we're going to need to plug one
and two into the function and then we'll talk about how this applies to the intermediate
value theorem so we're going to plug both one and two into our function so we'll say
f of one is one to fourth plus one minus three which of course will just give us one plus
one minus three which will give us negative one then if we plug in two the other end of
our interval f of two we'll get two to the fourth plus two minus three which will give
us sixteen plus two minus three which of course is just fifteen now without thinking about
it too much what we could assume is that since the function has a negative value on one end
of the interval at one and a positive value at the other end of the interval at two the
graph has to cross the x axis at some point in the interval and therefore the function
must have a root but we haven't really proven it and the only way we can prove it is with
the intermediate value theorem which basically tells us that if you take a point here a and
another point be along the x axis and you draw a graph let's say it looks like this
the graph of a function that a comes up to meet the graph at this point and if we go
over here to the left we can say that this point here is f of a whatever we get when
we plug a into our function gives us f of a over here same thing if we follow b up to
meet the graph like this and then we come over here until we meet y axis this point
here along the y axis is f of b what the intermediate value theorem tells us is that the function
must assume every single value between a and b so all of these values in here on this range
have to be on the graph of the function so what it says is if I name a point we'll call
it f of c a point in between f of a and f of b it has to be in between these two but
if it's a point on the y axis in between these two called f of c there must be there must
by the theorem there must be a point on the graph or a point along the x axis c that gives
me f of c so according to the intermediate value theorem if I have a and f of a and I
have b and f of c then since the function must assume every value in between f of a
and f of b there has to a point c here along the x axis that's between a and b that gives
me the value f of c so if we apply this to our problem what we see is that we have basically
the points here one and two and we now know that f of a is negative one and f of b is
fifteen so what this tells us is f of c our point here is zero the root right is zero
the intermediate value theorem tells us that there must a point c here in between a and
b where the function assumes the value of zero in other words f of c so this is how
we use the intermediate value theorem to show that there must be a root of the function
in between one and two and we can we can indicate that just by saying that f of one is less
than zero and zero of course is less than f of two and the fact that this inequality
is true means that there is a root of our function on the interval of one to two so
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