Read this section to learn how the first derivative is used to determine the shape of functions. Work through practice problems 1-9.
The First Derivative and the Shape of a Function f
This section examines some of the interplay between the shape of the graph of and the behavior of
. If we have a graph of
, we will see what we can conclude about the values of
.
If we know values of
, we will see what we can conclude about the graph of
.
Graphically, is increasing (decreasing) if, as we move from left to right along the graph of
, the height of the graph increases (decreases).
These same ideas make sense if we consider to be the height (in feet) of a rocket at time
seconds. We naturally say that the rocket is rising or that its height is increasing if the height
increases over a period of time, as
increases.
Example 1: List the intervals on which the function given in Fig. 1 is increasing or decreasing.
Fig. 1
Solution: is increasing on the intervals
and
.
is decreasing on
and
. On the interval
the function is not increasing or decreasing, it is constant. It is also valid to
say that
is increasing on the intervals
and
as well as many others, but we usually talk about the longest intervals on which
is monotonic.
Practice 1: List the intervals on which the function given in Fig. 2 is increasing or decreasing.
Fig. 2
If we have an accurate graph of a function, then it is relatively easy to determine where is monotonic, but if the function is defined by an equation, then a little more work is required. The next two theorems relate the values of the derivative
of
to the monotonicity of
. The first theorem says that if we know where
is monotonic, then we also know something about the values of
. The second theorem says that if we know about the values of
then we can draw conclusions about where
is monotonic.
First Shape Theorem
For a function which is differentiable on an interval
;
(i) if is increasing on
, then
for all
in
(ii) if is decreasing on
, then
for all
in
(iii) if is constant on
, then
for all
in
.
Proof: Most people find a picture such as Fig. 3 to be a convincing justification of this theorem: if the graph of increases near a point
, then the tangent line is also increasing, and the slope of the tangent line is positive (or
perhaps zero at a few places). A more precise proof, however, requires that we use the definitions of the derivative of
and of "increasing".
Fig. 3
(i) Assume that is increasing on
. We know that
is differentiable, so if
is any number in the interval
then
, and
this limit exists and is a finite value.
If is any small enough positive number so that
is also in the interval
, then
and
. We know that the numerator,
, and the denominator,
, are both positive
so the limiting value,
, must be positive or zero:
.
(ii) Assume that is decreasing on
: The proof of this part is very similar to part (i). If
, then
since
is decreasing on
. Then the numerator of the limit,
, will be negative
and the denominator,
, will still be positive, so the limiting value,
, must be negative or zero:
.
(iii) The derivative of a constant is zero, so if is constant on
then
for all
in
.
The previous theorem is easy to understand, but you need to pay attention to exactly what it says and what it does not say. It is possible for a differentiable function which is increasing on an interval to have horizontal tangent lines at some places in the interval (Fig 4).
Fig. 4
It is also possible for a continuous function which is increasing on an interval to have an undefined derivative at some places in the interval (Fig. 4).
Finally, it is possible for a function which is increasing on an interval to fail to be continuous at some places in the interval (Fig. 5).
Fig. 5
The First Shape Theorem has a natural interpretation in terms of the height and upward velocity
of a helicopter at time
. If the height of the helicopter is increasing (
is increasing ), then the helicopter has a positive or zero upward velocity:
. If the height of the helicopter is not changing, then its upward velocity is
.
Example 2: Fig. 6 shows the height of a helicopter during a period of time. Sketch the graph of the upward velocity of the helicopter, .
Fig. 6
Solution: The graph of is shown in Fig.7. Notice that the
has a local maximum when
and
, and
and
.
Similarly,
has a local minimum when
, and
. When
is increasing,
is positive. When
is decreasing,
is negative.
Fig. 7
Practice 2: Fig. 8 shows the population of rabbits on an island during 6 years. Sketch the graph of the rate of population change, , during those years.
Fig. 8
Example 3: The graph of is shown in Fig. 9. Sketch the graph of
.
Fig. 9
Solution: It is a good idea to look first for the points where or where
is not differentiable, the critical points of
. These locations are usually easy to spot, and they naturally break the problem into several smaller
pieces. The only numbers at which
are
and
, so the only places the graph of
will cross the
-axis are at
and
, and we can plot the point
and
on the graph
of
. The only place that
is not differentiable is at the "corner" above
, so the graph of
will not have a point for
. The rest of the graph of
is relatively easy:
if then
is decreasing so
is negative,
if then
is increasing so
is positive,
The graph of is shown in Fig. 10.
is continuous at
, but
is not differentiable at
, as is indicated by the "hole" in the graph.
Fig. 10
Practice 3: The graph of is shown in Fig. 11. Sketch the graph of
. (The graph of
has a "corner" at
.)
Fig. 11
The next theorem is almost the converse of the First Shape Theorem and explains the relationship between the values of the derivative and the graph of a function from a different perspective. It says that if we know something about the values of ,
then we can draw some conclusions about the shape of the graph of
.
Second Shape Theorem
For a function which is differentiable on an interval I;
(i) if for all
in the interval
, then
is increasing on
,
(ii) if for all
in the interval
, then
is decreasing on
,
(iii) if for all
in the interval
, then
is
constant on
.
Proof: This theorem follows directly from the Mean Value Theorem, and part (c) is just a restatement of the First Corollary of the Mean Value Theorem.
(a) Assume that for all
in
and pick any points a and
in
with
. Then, by the Mean Value Theorem, there is a point
between
and
so that
, and we can conclude that
and
. Since
implies that
, we know that
is increasing on
.
(b) Assume that for all
in
and pick any points
and
in
with
. Then there is a point
between
and
so that
, and we can conclude that
so
.
Since
implies that
, we know
is decreasing on
.
Practice 4: Rewrite the Second Shape Theorem as a statement about the height and upward velocity
of a helicopter at time
seconds.
The value of the function at a number tells us the height of the graph of
above or below the point
on the
-axis. The value of
at a number
tells us whether the graph of
is increasing or decreasing (or neither)
as the graph passes through the point
on the graph of
. If
is positive, it is possible for
to be positive, negative, zero or undefined: the value of
has absolutely
nothing to do with the value of
. Fig. 12 illustrates some of the combinations of values for
and
.
Fig. 12
Practice 5: Graph a continuous function which satisfies the conditions on and
given below
The Second Shape Theorem is particularly useful if we need to graph a function which is defined by an equation. Between any two consecutive critical numbers of
, the graph of
is monotonic (why?). If we can find
all of the critical numbers of
, then the domain of
will be naturally broken into a number of pieces on which
will be monotonic.
Example 4: Use information about the values of to help graph
.
Solution: so
only when
or
.
is a polynomial so it is always
defined. The only critical numbers for
are
and
, and they divide the real number line into three pieces on which
is monotonic:
and
.
If , then
(negative number)(negative number)
so
is increasing.
If , then
(positive number)(negative number)
so
is decreasing.
If , then
(positive number)(positive number)
so
is increasing.
Even though we don't know the value of anywhere yet, we do know a lot about the shape of the graph of
: as we move from left to right along the
-axis, the graph of
increases until
, then the graph decreases until
, and then the graph increases again (Fig. 13) The graph of
makes "turns" when
and
.
Fig. 13
To plot the graph of , we still need to evaluate
at a few values of
, but only at a very few values.
, and
is a local maximum of
.
, and
is a local minimum of
. The graph
of
is shown in Fig. 14.
Fig. 14
Practice 6: Use information about the values of to help graph
.
Example 5: Use the graph of in Fig. 15 to sketch the shape of the graph of
. Why isn't the graph of
enough to completely determine the graph of
?
Fig. 15
Solution: Several functions which have the derivative we want are given in Fig. 16 , and each of them is a correct answer. By the Second Corollary to the Mean Value Theorem, we know there is a whole family of parallel functions which have the derivative
we want, and each of these functions is a correct answer. If we had additional information about the function such as a point it went through, then only one member of the family would satisfy the extra condition and that function would be the only
correct answer.
Fig. 16
Practice 7: Use the graph of in Fig. 17 to sketch the shape of a graph of
.
Fig. 17
Practice 8: A weather balloon is released from the ground and sends back its upward velocity measurements (Fig. 18). Sketch a graph of the height of the balloon. When was the balloon highest?
Fig. 18
Source: Dale Hoffman, https://s3.amazonaws.com/saylordotorg-resources/wwwresources/site/wp-content/uploads/2012/12/MA005-4.3-First-Derivative.pdf This work is licensed under a Creative Commons Attribution 3.0 License.