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C
urve
S
ketching
Our goal is to understand how the graph of a function relates to the graph of its
derivative, and vice versa. In particular, we want to be able to pick out key characteristics
from each graph and use that information to sketch a graph of either the derivative or
antiderivative.
Armed with the power rule for polynomials which states that
()
1
n
d
ax
anx
dx
−
=
n
, let us
begin by looking at a particular function and its derivative.
Example 1
: Graph
f
(
x
) =
x
3
–
x
2
+
x
+ 1 and sketch its derivative.
Solution
: When sketching graphs of functions and derivatives, it is often useful to stack
them on top of each other, with the same
x
axis being used. That way, we can see how
attributes of one graph appear in the other.
Notice that if
f
(
x
) =
x
3
– 4
x
2
+ 2
x
+ 1, then we have that
f
′
(
x
) = 3
x
2
– 8
x
+ 2.

2

1
1
2
3
4
x

4

3

2

1
1
2
3
4
y
The graph of
f
(
x
) appears to the left with a
graph of
f
′
(
x
) below it. (We shall construct
f
′
(
x
) shortly, but for the time being, we can
use it to help illustrate some relationships.)
Graph of
f
(
x
)

2

1
1
2
3
4
x

4

3

2

1
1
2
3
4
y
Notice that
f
(
x
) has a local max. around
x
= 1/3 and a local min. around
x
= 7/3.
If we look at the graph of
f
′
(
x
), we notice
that it has
x
intercepts at precisely the
same points.
Graph of
f
′
(
x
)
This should not surprise us, since, by
definition, the slope of
f
(
x
) is zero at its
local max/min and the graph of
f
′
(
x
) is a
graph of the slopes of
f
(
x
).
Thus, the points where f(x) has zero slope
corresponds to xintercepts on f
′
(x)
.
Figure 1:
Graphs of
f
(
x
) and
f
′
(
x
)
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View Full DocumentKnowing the
x
intercepts is one piece of the graph. The question remains: what does the
graph do between the
x
intercepts?
Recall, one interpretation of the derivative is the slope of the tangent line at a point.
Using that property, we can determine the intervals where
f
′
(
x
) is positive and where
f
′
(
x
)
is negative.
Looking back at our graph of
f
(
x
), we see that the function
f
(
x
) is increasing (i.e.
f
(
x
) has
positive slope, i.e.
f
′
(
x
) is positive) on the intervals (
¶
, 1/3) and (7/3,
¶
). Similarly, we
see that
f
(
x
) is decreasing (i.e.
f
(
x
) has negative slope, i.e.
f
′
(
x
) is negative) on the interval
(1/3, 7/3).
So, we know that
f
′
(
x
) lies above the
x
axis to the left of the first
x
intercept and to the
right of the second
x
intercept; in between the
x
intercepts,
f
′
(
x
) lies below the
x
axis.
So, let us make a crude sketch of what
f
′
(
x
) should look like, given the information we
have found thus far. (Compare this to the actual graph above.)

2

y
4
1
1
2
3
4
x

4

3

2

1
1
2
3
Figure 2:
A rough sketch of
f
′
(
x
)
When we sketch our graph, we need to follow the general feel of the graph that is given.
What do we mean by that? Well, if
f
(
x
) appears to have nice, smooth curves, then we
should draw something similar for the graph of
f
′
(
x
). That is,
f
′
(
x
) should not be rough or
jagged, as is shown above, since the graph of
f
(
x
) does not look like this.
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 Fall '08
 staff
 Calculus, Derivative

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