Curve Sketch
Definition (Critical Points) We will call the number
a first order critical number if
or
does not exist and a second-order critical number if
or
does not exist.
Definition (Concave Up and Concave Down) If the graph of
lies above all of its tangents on an interval
it is called concave upward on
If the graph of
lies below all of these tangents, it is called concave downward on
Definition (Infection Point) A point
on a curve is called an inflection point of the graph is concave up on one side of
and concave down on the other side.
Proposition (Test for Concavity) Suppose
is twice differentiable on an interval
Then,
(i) If
for all
in
then the graph of
is concave upward on
![]()
(ii) If
for all
in
then the graph of
is concave downward on
![]()
Example (Concavity and Inflection Points) Determine where the curve
is concave upward, where it is concave downward, and where the points of inflection are (if any).
Solution. We use the Concavity Test and find the first and second derivatives:
![]()
![]()
which allows us to find the second order critical numbers, namely
when
and
We summarize the Concavity Test in the following table:
![curve sketch _gr_31.gif]](pages/curve-sketch/Images/curve-sketch_gr_31.gif)
Therefore,
is concave up on
and concave down on
The points
and
are inflection points.
Example (Concavity and Inflection Points) Determine where the curve
is concave upward and where it is concave downward. Find all inflection points, local extrema, and sketch the curve.
Solution. We apply the first derivative test and the concavity test by find the first and second derivatives
![]()
![]()
and then finding the first and second order critical numbers, namely:
![]()
![]()
Therefore, the first order critical numbers are
and the second order critical numbers are
We summarize the First Derivative Test and the Concavity Test in the following table:
![curve sketch _gr_45.gif]](pages/curve-sketch/Images/curve-sketch_gr_45.gif)
Therefore, the function has a local maximum at
and a local minimum at
This function
is increasing on the
and decreasing on
The point
is an inflection point because
is concave up on the interval
and concave down on
Definition (Critical Points) We will call the number
a first order critical number if
or
does not exist and a second-order critical number if
or
does not exist.
Proposition (Second Derivative Test) Suppose
in continuous on an open interval that contains
with
Then
(i) If
then
has a relative (local) minimum at
![]()
(ii) If
then
has a relative (local) maximum at
Warning. The second derivative test doe not tell us anything if both
and
For example, if
and
both
and
The point
is a minimum for
but is neither a maximum not a minimum for
However, the first derivative test is still useful.
Example (Second Derivative Test) Use the second derivative test to determine whether each critical number of the function
corresponds to a relative maximum, a relative minimum, or neither.
Solution. We find the first order critical numbers by
![]()
and so
and
are the critical numbers. Now to apply the Second Derivative Test we find the second derivative,
Since
the point
is a local maximum and since
the point
is a local minimum.
Example (Second Derivative Test) Use the second derivative test to determine whether each critical number of the function
corresponds to a relative maximum, a relative minimum, or neither.
Solution. We find the first order critical numbers by
![]()
and so
are the first order critical numbers of
Note that, even though
is undefined, so is
and so
is not a critical number. We compute
and since
the point
is a local minimum and since
the point
is a local maximum.
Example (Sketching the Graph of a Function) For the function
. Find all first and second order critical numbers. Apply the First Derivative Test, Concavity Test, and the Second Derivative Test. Sketch the graph of the function.
Solution. The first and second derivatives are:
![]()
![]()
Solving
we find the first order critical numbers to be
Solving
we find the second order critical numbers to be
Applying the First Derivative Test we find,
![curve sketch _gr_109.gif]](pages/curve-sketch/Images/curve-sketch_gr_109.gif)
Checking these local extrema by using the Second Derivative Test: determining
and so
is a local minimum and
and so
is a local maximum. Applying the Concavity Test:
![curve sketch _gr_114.gif]](pages/curve-sketch/Images/curve-sketch_gr_114.gif)
A sketch of the graph of
follows:
![curve sketch _gr_116.gif]](pages/curve-sketch/Images/curve-sketch_gr_116.gif)
Example (Sketching the Graph of a Function) For the function
. Find all first and second order critical numbers. Apply the First Derivative Test, Concavity Test, and the Second Derivative Test. Sketch the graph of the function.
Solution. The first and second derivatives are:
![]()
![]()
Note that
is not a critical number because
is not defined. Solving
we find the first order critical numbers to be
Solving
we find the second order critical numbers to be
Applying the First Derivative Test we find,
![curve sketch _gr_127.gif]](pages/curve-sketch/Images/curve-sketch_gr_127.gif)
Checking this local extrema by using the Second Derivative Test: determining
and so
is a local minimum.
Applying the Concavity Test:
![curve sketch _gr_130.gif]](pages/curve-sketch/Images/curve-sketch_gr_130.gif)
A sketch of the graph of
follows:
![curve sketch _gr_132.gif]](pages/curve-sketch/Images/curve-sketch_gr_132.gif)
There are four possibilities for unbounded behavior of a derivative
around a given real number
They are:
![curve sketch _gr_136.gif]](pages/curve-sketch/Images/curve-sketch_gr_136.gif)
which are called vertical tangents; and when these limits differ in sign they are:
![curve sketch _gr_137.gif]](pages/curve-sketch/Images/curve-sketch_gr_137.gif)
which are called cusps. Here are the formal definitions of vertical tangents and cusps followed by an example of each.
Definition (Vertical Tangent) Suppose the function
is continuous at the point
Then the graph of
has a vertical tangent at
if one of the following holds:
Example (Vertical Tangent) Sketch the graph of
and explain why there is a vertical tangent at
.
Solution. Using the product rule we find the derivative as
![]()
![]()
![]()
![]()
To determine any vertical tangents we consider where
is undefined. Notice that at
the derivative is undefined but
Thus, the point
is a candidate for a being a vertical tangent to the graph of
We check the following limits to determine if
is a vertical tangent.
Since
and
as
![]()
Since
and
as
![]()
By the definition of a vertical tangent
is a vertical tangent which can be seen from the sketch of the graph of
![curve sketch _gr_166.gif]](pages/curve-sketch/Images/curve-sketch_gr_166.gif)
Definition (Cusp) Suppose the function
is continuous at the point
Then the graph of
has a cusp at
if one of the following holds:
Example (Cusp) Sketch the graph of
and explain why there is a cusp at
.
Solution. Using the product rule we find the derivative as
![]()
![]()
![]()
![]()
![]()
To determine any cusps we consider where
is undefined. Notice that at
the derivative is undefined but
Thus, the point
is a candidate for a being a cusp for the graph of the function
. We check the following limits to determine if
is a cusp.
Since
and
as
![]()
Since
and
as
![]()
By the definition of a cusp
is a cusp which can be seen from the sketch of the graph of
![curve sketch _gr_197.gif]](pages/curve-sketch/Images/curve-sketch_gr_197.gif)
Example (Curve Sketching) Sketch the graph of the rational function
![]()
showing all special features.
Solution. The domain is
![]()
The
and
intercepts are both
Since
the function is even. The curve is symmetric about the
-axis.
![curve sketch _gr_206.gif]](pages/curve-sketch/Images/curve-sketch_gr_206.gif)
Therefore, the line
is a horizontal asymptote. Since the denominator is 0 when
we compute the following limits:
![curve sketch _gr_209.gif]](pages/curve-sketch/Images/curve-sketch_gr_209.gif)
Therefore, the lines
and
are vertical asymptotes.
Next we find the derivative function.
![]()
Since
when
and
when
is increasing on
and
and decreasing on
and
The only critical number is
Since
changes sign from positive to negative at
is a local maximum by the First Derivative Test. Also,
![]()
Since
for all
we have
and
Thus the curve is concave downward on the intervals
and
and concave downward on
There is no point of inflection since
and
are not in the domain of
Here is a sketch of the graph:
![curve sketch _gr_239.gif]](pages/curve-sketch/Images/curve-sketch_gr_239.gif)
Example (Curve Sketching) Sketch the graph of the trigonometric function
![]()
showing all special features.
Solution. The domain is
The
-intercept is
The
-intercept occur when
![]()
which is precisely when
and
, because we need only consider
since function is periodic via,
![]()
There are no asymptotes. Computing the derivative,
![]()
![]()
![]()
Thus,
when
or when
so in
when consider the critical number
and
and
Applying the First Derivative Test we find:
![curve sketch _gr_261.gif]](pages/curve-sketch/Images/curve-sketch_gr_261.gif)
Computing
![]()
we find the second order critical numbers as
and
where
and
Applying the Concavity Test we find,
![curve sketch _gr_268.gif]](pages/curve-sketch/Images/curve-sketch_gr_268.gif)
Here is a sketch of the graph:
![curve sketch _gr_269.gif]](pages/curve-sketch/Images/curve-sketch_gr_269.gif)
![]()
Curve Sketch
Published by Library of Math -- Online math organized by subject into topics.
Written by Smith, David A.
http://www.libraryofmath.com/curve-sketch.html


