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119
Lecture 16
Curve Sketching
Consider the function f whose graph appears in Figure 1, and think about the slopes of the
tangent lines to the curve.
a
b
c
f
Figure 1
It should be fairly obvious that the lines tangent to the curve at x = b and x = c are horizontal
lines with a slope of zero. It follows, then, that b and c are real roots of f ' . It should be equally
obvious that the tangent lines on the interval ( a, b ) have negative slopes as do the lines tangent
to f at points on the interval ( c, ∞ ) . From this we gather that f ' < 0 on ( a, b ) ∪ ( c, ∞ ) . Finally,
it is clear that lines tangent to f on the intervals ( −∞, a ) ∪ ( b, c ) all have positive slopes, which
means f ' > 0 over the same intervals. All this information is summarized by the graph of f '
below in Figure 2.
a
b
c
f'
Figure 2
Notice that where f ' > 0 , f is increasing and where f ' < 0 , f is decreasing. We can
generalize this statement, and we will.
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Lecture 16
The Behavior Theorem: If f ' ( x ) > 0 on an interval, then f is increasing on that
interval, and if f ' ( x ) < 0 on an interval, then f is decreasing on that interval.
Returning to our discussion of Figure 1 and Figure 2. Notice that where f ' = 0 or is
undefined there occurs a change in behavior in f from increasing to decreasing or vice versa. This
statement is not always true. Imagine that f was defined in such a way that the graph of f in
Figure 1 was reflected over the x-axis over the interval ( −∞, a ) , but the remainder of the graph
remained as is. In such a case, the discontinuity at a would not correspond to a change from one
behavior to another since f would be decreasing on both sides of a. Another exception would
occur on a curve like g shown below at x = n where g' would equal zero but would not
correspond to a change in behavior in g since g increases on either side of n.
n
Figure 3
Nevertheless, points where f ' = 0 or is undefined correspond to locations where f could change
behavior. Moreover, if these points occur in the domain of f, they could correspond to a "peak"
or "turn-around" in f, for this reason we will define the points along the x-axis where f ' = 0 or is
undefined as critical numbers of f if they occur in the domain of f.
A critical number of function f is a number c in the domain of f such that either
f ' ( c ) = 0 of f ' ( c ) does not exist.
Remember that critical numbers are x-values where f might have a "peak" or "turn-around." We
want to make use of this fact. To do so, we first need a definition.
If there exists an interval of values (a,b) containing c with a < c < b or b < c < a
over which f is continuous and such that f ( c ) ≥ f ( x ) or f ( c ) ≤ f ( x ) for all x on
the interval, then f ( c ) is a local extremum, also called a relative extremum. If
f ( c ) ≥ f ( x ) for all x on [a,b], then f ( c ) is a local (or relative) maximum. If
f ( c ) ≤ f ( x ) for all x on [a,b], then f ( c ) is a local (or relative) minimum.
This definition gives us a term for the "turn-around points" on f. If the turn-around occurs at a
"peak," it is a local maximum (see Figure 4). If the turn-around occurs at a "valley," it is a local
minimum (see Figure 5).
Figure 5
c
Figure 4
c
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Lecture 16
With local extrema defined, we can proceed to Fermat's Theorem.
Fermat's Theorem: If f has a local maximum or local
minimum at c, then c is a critical number of f.
One interesting aspect of Fermat's Theorem is that the converse is not necessarily true. As we
stressed in the previous paragraph, if c is a critical number of f, then f may or may not have a
local maximum or local minimum at c. Nevertheless, Fermat's Theorem is a powerful tool for
sketching curves because it allows us to think in the converse then investigate.
For example, let's consider f ( x ) = x 5 + 2 x 4 + x 3 . Suppose we want to sketch the curve
of this polynomial. We are interested in its intercepts, behavior, and extrema. Determining the
intercepts is uninteresting, so we will just list them as (0, −1) and ( 0, 0 ) . To determine behavior
and to find extrema, we will find f ' as below.
f ( x ) = x5 + 2 x 4 + x3
f ' ( x ) = 5 x 4 + 8 x3 + 3x 2
Next, we will find the critical numbers of f—that is, the x-values such that f ' does not exist or is
equal to zero. Since f ' is a polynomial and defined for all x-values, we conclude that the roots
of f ' are the only critical numbers of f.
5 x 4 + 8 x3 + 3x 2 = 0
x 2 ( 5 x 2 + 8 x + 3) = 0
xx ( 5 x + 3)( x + 1) = 0
3
x = 0, x = − , x = −1
5
Thinking of the converse of Fermat's Theorem, we suspect that these three critical
numbers correspond to extrema on f, so we investigate by plotting the three numbers on a
number line and evaluating f ' for values left and right of each one, making conclusions about f
as we go. Because we have Rolle's Theorem and all the roots of f ' , it is only necessary to
choose one value for each interval left of, right of, or between critical numbers, and any one
number on each interval will tell us the sign of f ' for every x-value on the interval.
⎛ 4⎞
f '⎜ − ⎟ < 0
⎝ 5⎠
f ' ( −2 ) > 0
f increases
−1
f decreases −
⎛ 1⎞
f '⎜ − ⎟ > 0
⎝ 5⎠
3
5
f increases
f ' (1) > 0
0
f increases
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Lecture 16
Applying the same logic that supports Rolle's Theorem to f ' , we know that f ' is positive on
the intervals ( −∞, −1) ∪ ( −3 5, 0 ) ∪ ( 0, ∞ ) . Correspondingly, by the Behavior Theorem, we
conclude that f increases on the same interval ( −∞, −1) ∪ ( −3 5, 0 ) ∪ ( 0, ∞ ) . Similarly, we know
f ' is negative on the interval ( −1, −3 5 ) and f decreases along ( −1, −3 5 ) . Note, that f does not
change its behavior at the critical number zero, which shows the converse of Fermat's Theorem
is not true for all critical numbers. At –1 and at −3 5 , however, the function does change
behavior, which indicates that f ( −1) and f ( − 3 5 ) are local extrema. Selecting an arbitrary
interval [a,b] such that –1 greater than one endpoint but less than the other and comparing
f ( −1) to the value of the points near it, we can determine if f ( −1) is a minimum or maximum.
We do the same for –3/5. Or, we can employ our number line diagram. If we superimpose a
sketch of general behavior that matches our conclusions, it becomes apparent that f ( −1) must
be a local maximum while f ( − 3 5 ) must be a local minimum.
⎛ 4⎞
f '⎜ − ⎟ < 0
⎝ 5⎠
f ' ( −2 ) > 0
f increases
−1
f decreases −
⎛ 1⎞
f '⎜ − ⎟ > 0
⎝ 5⎠
3
5
f increases
f ' (1) > 0
0
f increases
The process of testing the sign of f ' on either side of the critical numbers of f is called the
First Derivative Test.
To identify the actual extrema-values, we simply evaluate f ( −1) and f ( − 3 5 ) . Since
f ( − 3 5 ) = −0.03456 , we know –0.03456 is a local minimum that occurs when x = –3/5.
Since f ( −1) = 0 , we know zero is a local maximum that occurs when x = –1. Plotting the local
extrema and intercepts and heeding the intervals of increasing and decreasing behavior, we can
sketch the curve with fair accuracy.
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Lecture 16
Practice Problems in Calculus: Concepts and Contexts by James Stewart
1st ed. problem set:
Section 2.10 #1, #11a, #11b, #19, #20; Section 4.2 #25–35 odd;
& Section 4.3 #7–11 odd parts a & b, #13–19 odd parts a & b
Section 2.10 #1, #11a, #11b, #21, #22, #23a; Section 4.2 #23–33 odd;
& Section 4.3 #7–13 odd parts a & b, #17–23 odd parts a & b
Section 2.9 #1, #11a, #11b, #21, #22, #23a; Section 4.2 #23–35 odd;
& Section 4.3 #7–13 odd parts a & b, #19–27 odd parts a & b
1st ed. problem set:
3rd ed. problem set:
Practice Problems in Calculus: Early Transcendentals by Briggs and Cochran
1st ed. problem set:
Section 4.1 #7, #23-29 odd
Section 4.2 #11, #15, #17, #25-29 odd, #31-37 odd parts a & b
Possible Exam Problems
#1 State the Behavior Theorem.
Answer:
The Behavior Theorem: If f ' ( x ) > 0 on an interval, then f is increasing on that
interval, and if f ' ( x ) < 0 on an interval, then f is decreasing on that interval.
#2 State the definition of a critical number of f.
Answer:
A critical number of function f is a number c in the domain of f such that either
f ' ( c ) = 0 or f ' ( c ) does not exist.
3
8
#3 Given g ( x ) = 4 x 5 − x 5 . Find the critical numbers of g.
Answer:
0 and 1.5
#4 Given the graph of H ' below, what conclusions can we make regarding H. Assume the
scales of the x and y axes is 1 unit per tick mark.
Answer:
H ( 2 ) is a local maximum. H increases along ( −∞, 2 ) & decreases along ( 2, ∞ ) .
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Lecture 16
Example Exercise 1
Consider the function f ( x ) = x
2
3
( x + 1)
3
. Find the critical numbers of the function.
Critical numbers of f are numbers in the domain of f that are not in the domain of f ' or
that are roots of f ' . We start by finding f ' .
2
f ( x ) = x 3 ( x + 1)
3
2
⎤
d ⎡
3
3 d ⎡
⎤
f '( x) = x ⋅
x + 1) + ( x + 1) ⋅ ⎢ x 3 ⎥
(
⎦
dx ⎣
dx ⎣ ⎦
2
3
2
2
−1
d
3 2
[ x + 1] + ( x + 1) ⋅ x 3
dx
3
2
1
−
2
3 2
f ' ( x ) = x 3 ⋅ 3 ( x + 1) ⋅1 + ( x + 1) ⋅ x 3
3
2
1
2 −
2
3
f ' ( x ) = 3 x 3 ( x + 1) + x 3 ( x + 1)
3
1
1 −
2
f ' ( x ) = x 3 ( x + 1) ⎡⎣9 x + 2 ( x + 1) ⎤⎦
3
f ' ( x ) = x 3 ⋅ 3 ( x + 1) ⋅
f '( x)
2
( x + 1) (11x + 2 )
=
2
33 x
Since the domain of f includes all real numbers, any real number that is a root of f ' or not in
the domain of f ' will be a critical number of f by definition.
The roots of the denominator, 3 3 x , represent numbers not in the domain of f ' . Thus, zero
is a critical number of f . The roots of the numerator, ( x + 1) (11x + 2 ) , represent roots of f ' .
2
Thus, −1 and − 2 11 are also critical numbers of f .
The critical numbers of f are −1, − 2 11, and 0 .
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Lecture 16
Example Exercise 2
Show that y = ln ( x − 1) always increases.
To apply the Behavior Theorem, find dy dx .
y = ln ( x − 1)
y'=
1
x −1
The domain of y = ln ( x − 1) only includes x-values such that x > 1 . Clearly, y ' > 0 ∀ x > 1 .
Hence, the function increases throughout its domain by the Behavior Theorem.
Example Exercise 3
Find any local extrema of f ( x ) = sin ( x ) ⋅ cos ( x ) along ( 0, π ) .
To apply the Behavior Theorem, find f ' .
f ( x ) = sin ( x ) ⋅ cos ( x )
d
d
⎡⎣cos ( x ) ⎤⎦ + cos ( x ) ⋅ ⎡⎣sin ( x ) ⎤⎦
dx
dx
2
2
f ' ( x ) = − sin ( x ) + cos ( x )
f ' ( x ) = sin ( x ) ⋅
Since the derivative is defined for all real numbers, set f ' equal to zero, to find the critical
numbers of f. Remember that we are only interested in values on the interval ( 0, π ) .
− sin 2 ( x ) + cos 2 ( x ) = 0
cos 2 ( x ) = sin 2 ( x )
x=
π 3π
,
4 4
Note that f ' (π 6 ) > 0 , f ' (π 2 ) < 0 , and f ' ( 5π 6 ) > 0 . By the Behavior Theorem, the
⎛ π ⎞ ⎛ 3π ⎞
⎛ π 3π ⎞
function f ( x ) increases along ⎜ 0, ⎟ ∪ ⎜ , π ⎟ and decreases along ⎜ , ⎟ . Hence, a local
⎝ 4⎠ ⎝ 4
⎠
⎝4 4 ⎠
extrema occur when x = π 4 and when x = 3 π 4 .
1
⎛π ⎞ 1
⎛ 3π ⎞
The point f ⎜ ⎟ = is a local maximum. The point f ⎜ ⎟ = − is a local minimum.
2
⎝ 4 ⎠
⎝4⎠ 2
126
Application Exercise
Let x represent the parts per million of carbon monoxide in the ambient
atmosphere (ranging from 500 to 800). Suppose the function T ( x ) gives the
approximate duration in hours of human exposure before victims will fall into a
carbon monoxide induced coma.
T ( x ) = 0.0002 x 2 − 0.316 x + 127.9
Consider the domain of T ( x ) , which is 500 ≤ x ≤ 800 . Over what interval of the
domain does the time before a person succumbs to carbon monoxide poisoning
decrease?