The Geometry of Graphs
Difference Equations
to
Differential Equations
Section 3.9
The Geometry of Graphs
In Section 2.1 we discussed the graph of a function y = f (x) in terms of plotting points
(x, f (x)) for many different values of x and connecting the resulting points with straight
lines. This is a standard procedure when using a computer and, if the function is well
behaved and sufficiently many points are plotted, will produce a reasonable picture of the
graph. However, as we noted at that time, this method assumes that the behavior of the
graph between any two successive points is approximated well by a straight line. With a
sufficient number of points and a differentiable function, this assumption will be reasonable.
Yet to understand a graph fully, it is important to have alternative techniques to verify
the picture at least qualitatively. We have already developed several important aids for
understanding the shape of a graph, including techniques for determining the location of
local extreme values and techniques for finding intervals where the function is increasing
and intervals where it is decreasing. In this section we will use this information, along
with additional information contained in the second derivative, to piece together a picture
of the graph of a given function.
2
To see
√ the importance of the second derivative, consider the graphs of f (x) = x and
g(x) = x on the interval (0, ∞). Now
f 0 (x) = 2x
and
1
g 0 (x) = √ ,
2 x
5
4
3
2
1
1
2
3
4
Figure 3.9.1 Graphs of y = x2 and y =
1
5
√
x
Copyright c by Dan Sloughter 2000
2
The Geometry of Graphs
Section 3.9
1
5
4
-3
-2
-3
-2
1
-1
3
-1
2
-2
1
-3
1
-1
-1
2
3
2
3
-4
-5
Figure 3.9.2 Graphs of y = x2 and y = −x2 on (−∞, ∞)
so f 0 (x) > 0 and g 0 (x) > 0 for all x in (0, ∞). Thus f and g are both increasing on (0, ∞).
However, the graphs of f and g, as shown in Figure 3.9.1, are dramatically different. The
graph of f is not only increasing, but is becoming steeper and steeper as x increases,
whereas the graph of g is increasing, but flattening out as x increases. In other words, f 0
is itself an increasing function, causing the rate of growth of the function to increase with
x, while g 0 is a decreasing function, resulting in a decrease in the rate of growth of g and
a flattening out of the graph. In the terminology of the next definition, we say that the
graph of f is concave up on (0, ∞) and the graph of g is concave down on (0, ∞).
Definition Suppose f is differentiable on the open interval (a, b). If f 0 is an increasing
function on (a, b), then we say the graph of f is concave up on (a, b). If f 0 is a decreasing
function on (a, b), then we say the graph of f is concave down on (a, b).
Of course, to check for the intervals where f 0 is increasing and the intervals where f 0 is
decreasing, we consider where f 00 , the derivative of f 0 , is positive and where it is negative.
Proposition Suppose f is twice differentiable on the interval (a, b). If f 00 (x) > 0 for all
x in (a, b), then the graph of f is concave up on (a, b); if f 00 (x) < 0 for all x in (a, b), then
the graph of f is concave down on (a, b).
Example Two basic examples to keep in mind are f (x) = x2 and g(x) = −x2 . Since
f 00 (x) = 2 > 0 and g 00 (x) = −2 < 0 for all values of x, the graph of f is concave up on
(−∞, ∞) and the graph of g is concave down on (−∞, ∞). See Figure 3.9.2.
Example Consider g(t) = t3 . Then g 00 (t) = 6t, so g 00 (t) < 0 when t < 0 and g 00 (t) > 0
when t > 0. Hence the graph of g is concave down on (−∞, 0) and concave up on (0, ∞).
Notice in Figure 3.9.3 how, even though g is increasing on (−∞, ∞), the change in concavity
at (0, 0) changes the shape of the graph.
Definition A point on the graph of a function f where the concavity changes from up
to down or from down to up is called an inflection point.
Example
In our previous example, (0, 0) is an inflection point for the graph of g(t) = t3 .
Section 3.9
The Geometry of Graphs
3
3
2
1
-3
-2
1
-1
2
3
-1
-2
-3
Figure 3.9.3 Graph of g(t) = t3
Example
Let f (x) =
1
. Then
x
1
f 0 (x) = − 2
x
and
f 00 (x) =
2
.
x3
Hence f 0 (x) < 0 on both (−∞, 0) and (0, ∞), while f 00 (x) < 0 when x < 0 and f 00 (x) > 0
when x > 0. Thus f is decreasing on both (−∞, 0) and (0, ∞), but the fact that the graph
is concave down on (−∞, 0) shows up in the way the steepness of the graph increases as x
approaches 0 from the right, while the fact that the graph is concave up on (0, ∞) shows
up in the way the graph flattens out as x increases toward ∞. See Figure 3.9.4. Also
note that, although the concavity of the graph of f changes, the graph does not have an
inflection point since f is not defined at 0.
4
2
-4
2
-2
4
-2
-4
Figure 3.9.4 Graph of f (x) =
1
x
4
The Geometry of Graphs
Section 3.9
Note that if (c, f (c)) is an inflection point on the graph of a function f , then either
f (c) = 0 or f 00 is not defined at c. However, the converse does not hold. For example, if
f (x) = x4 , then f 00 (0) = 0, even though f 00 (x) = 12x2 is positive for all x in both (−∞, 0)
and (0, ∞).
From the foregoing, it is clear that f 0 and f 00 provide enough information to obtain a
good understanding of the shape of the graph of f . Specifically, to sketch the graph of f ,
we use the first derivative to find (1) intervals where f is increasing, (2) intervals where f
is decreasing, and (3) locations of any local extreme values; we use the second derivative
to find (1) intervals where the graph of f is concave up, (2) intervals where the graph f
is concave down, and (3) any inflection points. Combining this information with a few
values of the function, the location of any asymptotes, and information on the behavior of
f (x) as x goes to −∞ and as x goes to ∞, we can piece together a qualitatively accurate
picture of the graph of f .
00
Example
Consider f (x) = 3x2 − x3 + 2. Then
f 0 (x) = 6x − 3x2 = 3x(2 − x),
so the critical points of f are 0 and 2. Since f 0 (−1) = −9 < 0, f 0 (1) = 3 > 0, and
f 0 (3) = −9 < 0, f is decreasing on the intervals (−∞, 0) and (2, ∞) and increasing on
(0, 2). Moreover, this shows that f has a local minimum of 2 at x = 0 and a local maximum
of 6 at x = 2.
Next, we have
f 00 (x) = 6 − 6x = 6(1 − x),
so f 00 (x) = 0 when x = 1. Now 1 − x > 0 when x < 1 and 1 − x < 0 when x > 1, so
f 00 (x) > 0 on (−∞, 1) and f 00 (x) < 0 on (1, ∞). Hence the graph of f is concave up on
(∞, 1)and concave down on (1, ∞), and (1, 4) is an inflection point.
Combining this information with the values f (−1) = 6, f (3) = 2,
2
3
3
lim f (x) = lim (3x − x + 2) = lim x
x→−∞
x→−∞
x→−∞
and
2
3
3
lim f (x) = lim (3x − x + 2) = lim x
x→∞
x→∞
x→∞
3
2
−1+ 3
x
x
3
2
−1+ 3
x
x
= ∞,
= −∞,
we can easily draw a graph which, even though we are only plotting five points (the two
local extreme values, the inflection point, and one point on each side of these points),
captures the shape of the graph of f very well. See Figure 3.9.5.
Example
Consider g(x) = 12x5 + 15x4 − 40x3 − 10. Then
g 0 (x) = 60x4 + 60x3 − 120x2 = 60x2 (x2 + x − 2) = 60x2 (x + 2)(x − 1),
implying that g has three critical points, namely, x = −2, x = 0, and x = 1. Now 60x2 ≥ 0
for all values of x; x + 2 < 0 when x < −2 and x + 2 > 0 when x > −2; and x − 1 < 0
Section 3.9
The Geometry of Graphs
5
10
7.5
5
2.5
2
-2
4
-2.5
Figure 3.9.5 Graph of f (x) = 3x2 − x3 + 2
when x < 1 and x − 1 > 0 when x > 1. Thus g 0 (x) > 0 when x < −2 and when x > 1,
and g 0 (x) < 0 when −2 < x < 0 and when 0 < x < 1. So g is increasing on (−∞, −2) and
(1, ∞), and g is decreasing on (−2, 0) and (0, 1). In particular, g has a local maximum
of 166 at x = −2 and a local minimum of −23 at x = 1. Although g has neither a local
maximum nor a local minimum at the critical point 0, for drawing the graph of g it is
important to note that the slope of the curve is 0 at (0, −10).
Next,
g 00 (x) = 240x3 + 180x2 − 240x = 60x(4x2 + 3x − 4),
so g 00 (x) = 0 when x = 0 and when x2 + 3x − 4 = 0. Using the quadratic formula, the
latter equation has solutions
√
−3 − 73
x=
= −1.4430
8
and
√
−3 + 73
x=
= 0.6930,
8
rounding to four decimal places. Now 4x2 + 3x − 4 < 0 only when x is between the two
roots −1.4430 and 0.6930. Since 60x < 0 when x < 0 and 60x > 0 when x > 0, we
may conclude that g 00 (x) < 0 for x < −1.4430 and 0 < x < 0.6930, and g 00 (x) > 0 for
−1.4430 < x < 0 and x > 0.6930. Hence the graph of g is concave down on (−∞, −1.4430)
and (0, 0.6930) and concave up on (−1.4430, 0) and (0.6930, ∞). In particular, g has three
inflection points: (−1.4430, 100.1459), (0, −10), and (0.6930, −17.9349).
Adding to this information the values g(−3) = −631, g(2) = 294,
lim g(x) = lim (12x5 + 15x4 − 40x3 − 10)
x→−∞
15 40 10
5
− 2− 5
= lim x 12 +
x→−∞
x
x
x
x→−∞
= −∞,
6
The Geometry of Graphs
Section 3.9
400
300
200
100
-4
-3
-2
1
-1
2
3
-100
-200
-300
Figure 3.9.6 Graph of g(x) = 12x5 + 15x4 − 40x3 − 10
and
lim g(x) = lim (12x5 + 15x4 − 40x3 − 10)
x→∞
15 40 10
5
= lim x 12 +
− 2− 5
x→∞
x
x
x
x→∞
= ∞,
we can now sketch the graph of g. See Figure 3.9.6.
Example
For our final example, consider
h(t) =
Then
h0 (t) =
t2
.
t2 − 1
2t
(t2 − 1)(2t) − (t2 )(2t)
=− 2
,
2
2
(t − 1)
(t − 1)2
so h0 (t) = 0 when 2t = 0. Thus h has one critical point, t = 0. However, we must also
take into consideration the two points where h and h0 are not defined, namely, t = −1 and
t = 1. Now (t2 − 1)2 ≥ 0 for all t, so the sign of h0 is determined by the sign of −2t. Thus
h0 (t) > 0 when t < −1 and when −1 < t < 0, and h0 (t) < 0 when 0 < t < 1 and when
t > 1. In other words, h is increasing on (−∞, −1) and (−1, 0), and h is decreasing on
(0, 1) and (1, ∞). From this we see that h has a local maximum of 0 at t = 0. For the
second derivative, we have
h00 (t) =
−2(t2 − 1) + 8t2
6t2 + 2
(t2 − 1)2 (−2) − (−2t)(2(t2 − 1)(2t))
=
=
.
(t2 − 1)4
(t2 − 1)3
(t2 − 1)3
Since 6t2 + 2 > 0 for all values of t, it follows that h00 (t) 6= 0 for all t. However, as with the
first derivative, we need to consider the points t = −1 and t = 1 where h00 is not defined.
Now t2 − 1 < 0 only when −1 < t < 1, so h00 (t) < 0 when −1 < t < 1 and h00 (t) > 0 when
Section 3.9
The Geometry of Graphs
7
8
6
4
2
-3
-2
1
-1
2
3
-2
-4
-6
-8
Figure 3.9.7 Graph of h(t) =
t2
t2 − 1
t < −1 and when t > 1. Hence the graph of h is concave down on (−1, 1) and concave up
on (−∞, −1) and (1, ∞). Note, however, that there are no points of inflection.
Since h is not defined at t = −1 and t = 1, we need to check for vertical asymptotes
at these points. We have
t2
= ∞,
t→−1− t2 − 1
lim h(t) = lim
t→−1−
t2
= −∞,
lim + h(t) = lim + 2
t→−1
t→−1 t − 1
t2
= −∞,
t→1− t2 − 1
lim h(t) = lim
t→1−
and
t2
lim h(t) = lim 2
= ∞,
t→1+
t→1+ t − 1
showing that the graph of h has vertical asymptotes at t = −1 and t = 1. Finally,
t2
= lim
t→−∞ t2 − 1
t→−∞
1
lim h(t) = lim
t→−∞
and
t2
= lim
t→∞ t2 − 1
t→∞
lim h(t) = lim
t→∞
1−
1
1−
1
t2
1
t2
=1
=1
show that the graph of h has a horizontal asymptote at y = 1. With all of this geometric
information, we may now draw the graph of h, as shown in Figure 3.9.7.
8
The Geometry of Graphs
Section 3.9
Problems
1. Discuss the geometry of the graphs of each of the following functions. That is, find
the intervals where the function is increasing and where it is decreasing, find the
intervals where the graph is concave up and where it is concave down, find all local
extreme values and where they are located, find all inflection points, find any vertical
or horizontal asymptotes, and use this information to sketch the graph.
(a) f (x) = x2 − x
(b) g(t) = 3t2 + 2t − 6
(c) g(x) = x3 + 3x2
(d) f (t) = t4 + 2t2
(e) f (x) = x3 − 3x
(f) g(x) = 3x5 − 5x3
(g) h(x) = x5 − x3
1
(i) g(z) =
z−1
(h) f (x) = 3x5 − 5x4
1
(j) y(t) = 2
t +1
t
(l) h(t) =
1 + t2
x
(n) g(x) =
1 + 3x2
1
(p) f (x) = 2
x −1
z2
(r) f (z) = 2
z −4
(k) f (x) = x4 − 2x3
(m) h(t) =
t
t2 − 4
x2
1 + x2
2t + 1
(q) x(t) =
t−1
(o) f (x) =
2. Suppose the function f has the following properties:
f (0) = 0
f 0 (x) > 0 for x in (−∞, 2)
f 0 (x) < 0 for x in (2, ∞)
f 00 (x) < 0 for x in (−2, 6)
f 00 (x) > 0 for x in (−∞, −2) and for x (6, ∞)
lim f (x) = −2
x→−∞
lim f (x) = 0
x→∞
Sketch the graph of a function satisfying these conditions.
3. Suppose f (0) = 0 and f 0 (x) = x2 − 1.
(a) Sketch what the graph of f must look like.
(b) Graph f 0 on the same axes with f .
(c) Is there more than one function f which satisfies these conditions?
Section 3.9
The Geometry of Graphs
9
4. Suppose f (0) = 0 and f 0 (x) = x3 + x2 − 6x.
(a) Sketch what the graph of f must look like.
(b) Graph f 0 on the same axes with f .
(c) Is there more than one function f which satisfies these conditions?
1
.
t
(a) Sketch what the graph of g must look like on (0, ∞).
(b) Graph g 0 on the same axes with g.
(c) Is there more than one function g which satisfies these conditions?
5. Suppose g(1) = 0 and g 0 (t) =
6. Suppose f (0) = 1 and f 0 (x) = f (x). What must the graph of f look like? Is this
enough information to determine the graph of f?