- [Tutor] Let g of x equal one over x.
Can we use the mean value theorem
to say that the equation g prime of x is equal to one half
has a solution where negative one
is less than x is less than two,
if so, write a justification.
Alright, pause this video
and see if you can figure that out.
So the key to using the mean value theorem,
even before you even think about using it,
you have to make sure
that you are continuous over the closed interval
and differentiable over the open interval,
so this is the open interval here
and then the closed interval would include the end points.
But you might immediately realize
that both of these intervals contain x equals zero
and if x equals zero, the function is undefined
and if it's undefined there,
well it's not going to continuous
or differentiable at that point
and so no,
not continuous
or differentiable,
differentiable
over the interval,
over the interval.
Alright, let's do the second part.
Can we use the mean value theorem
to say that there is a value c such that g prime of c
is equal to negative one half
and one is less than c is less than two
if so, write a justification.
So pause the video again.
Alright, so in this situation, between one and two
on both the open and the closed intervals,
well, this is a rational function
and a rational function is going to be continuous
and differentiable at every point in its domain
and its domain completely contains
this open and closed interval
or another way to think about it,
every point on this open interval
and on the closed interval is in the domain,
so we can write g of x
is a rational
function,
which lets us know that it is continuous
and differentiable at every point in its domain,
at every point
in its domain,
the closed interval from one to two
is in domain
and so now, let's see what the average rate of change is
from one to two
and so we get g of two
minus g of one
over two minus one is equal to one half
minus one over one, which is equal to negative one half,
therefore,
therefore
by the
mean value theorem,
there must be a c
where one is less than c is less than two
and g prime of c is equal to
the average rate of change between the end points,
negative one half
and we're done, so we can put a big yes right over there
and then this is our justification.