Suppose that x and are in . We then have,
Now, using Property 5 of the
Integral Properties we can rewrite the first integral and then do a little
simplification as follows.
that and we get,
assume that and since we are still assuming that are in we know that
is continuous on and so by the Extreme
Value Theorem we know that there are numbers c and d in so that is the absolute minimum of in and that is the absolute maximum of in .
So, by Property 10 of the
Integral Properties we then know that we have,
Now divide both sides of this by h to get,
and then use (1)
Next, if we can go through the same argument above
except we’ll be working on to arrive at exactly the same inequality above. In other words, (2)
is true provided .
Now, if we take we also have and because both c and d are between x and . This means that we have the following two
The Squeeze Theorem then
tells us that,
but the left
side of this is exactly the definition of the derivative of and so we get that,
So, we’ve shown
that is differentiable on .
Now, the Theorem
at the end of the Definition of the Derivative section tells us that is also continuous on . Finally, if we take or we can go through a similar argument we used
to get (3)
using one-sided limits to get the same result and so the theorem at the end
of the Definition of the Derivative section will also tell us that is continuous at or and so in fact is also continuous on .