We want to show that for any point the interval
coming from any small interval
centred on
is larger than the interval
coming from the interval
of the same width as
, but centred on
(see the figure).

Figure 7: The interval J1 is smaller than the interval J2.
Since is the only point with
we know that
To match with our figure we’ll assume that
An analogous argument works when
Now choose a positive number with
(you’ll see later why we make this choice). From the definition of a derivative we know that there exists
such that for all
with
we have
![]() |
(1) |
Similarly, there exists such that for all
with
we have
![]() |
(2) |
Now choose any less than both
and
. Let
be the interval
and let
be the interval
Since is the only value for which
we know that the maximum of
on
must occur at one of the end-points of
. Call that point
. This means that the width of the interval
is
Since is the only value for which
we also know that the slope of
is negative throughout the interval
. On
the function
achieves its maximum at the lower end point
of
and its minimum at the upper end point
The width of the interval
is therefore
Now since both and
lie within
of
, we know from inequality 2 that
![]() |
and
![]() |
(The inequalities have changed direction because, since , we have multiplied by a negative number in each case.)
Putting this together gives
![]() |
(3) |
Since lies within
of
, we know from inequality 1 that
![]() |
(4) |
And since we have chosen so that
, putting inequalities 3 and 4 together gives
![]() |
Thus, interval is wider than interval
. QED.