Let a be any
number in the interval I. The tangent line to at is,
To show that is concave up on I then we need to show that for any x, ,
in I that,
or in other words, the tangent line is always below the
graph of on I. Note that we require because at that point we know that since we are talking about the tangent line.
Let’s start the proof off by first assuming that . Using the Mean Value Theorem on means there is a number c such that and,
With some rewriting this is,
(2)
Next, let’s use the fact that for every x on I. This means that the first derivative, ,
must be increasing (because its derivative, ,
is positive). Now, we know from the
Mean Value Theorem that and so because is increasing we must have,
(3)
Recall as well that we are assuming and so . If we now multiply (3)
by (which is positive and so the inequality
stays the same) we get,
Next, add to both sides of this to get,
However, by (2),
the right side of this is nothing more than and so we have,
but this is exactly what we wanted to show.
So, provided the tangent line is in fact below the graph
of .
We now need to assume . Using the Mean Value Theorem on means there is a number c such that and,
If we multiply both sides of this by 1
and then adding to both sides and we again arise at (2).
Now, from the Mean Value Theorem we know that and because for every x on I we know that the
derivative is still increasing and so we have,
Let’s now multiply this by ,
which is now a negative number since . This gives,
Notice that we had to switch the direction of the
inequality since we were multiplying by a negative number. If we now add to both sides of this and then substitute (2)
into the results we arrive at,
So, again we’ve shown that the tangent line is always
below the graph of .
We’ve now shown that if x is any number in I,
with the tangent lines are always below the graph
of on I
and so is concave up on I.
