CURVE SKETCHING
To sketch a curve, the first thing to do is determine where the function is increasing, decreasing,
or constant.
A function is increasing if, as x moves to the right, its graph moves up. This will happen when
the derivative of the function is positive.
A function is decreasing if, as x moves to the right, its graph moves down. This will happen
when the derivative of the function is negative.
A function is constant if, as x moves to the right, its graph neither rises nor falls. This will
happen when the derivative of the function is zero.
A positive derivative implies that the function is increasing; a negative derivative implies that the
function is decreasing; and a zero derivative on an entire interval implies that the function is
constant on that interval.
Example 1
Find the intervals over which the function y  x3 
3 2
x is increasing, decreasing, and constant.
2
2
1.5
1
0.5
2
1
1
0.5
1
1.5
2
2
3
Guidelines for finding intervals on which a function is increasing or decreasing
1. Locate the critical values (the places where the derivative is zero or does not exist), and
use these numbers to determine test intervals.
2. Determine the sign of f   x  at one test value in each interval.
3. Use the information in step 2 to determine if the function is increasing, decreasing, or
constant in that interval.
After you find the intervals over which a function is increasing or decreasing, it is not difficult to
locate the relative extrema of a function. The First Derivative Test is used to perform this task.
The First Derivative Test
The First Derivative Test is used to determine whether the function has a relative minimum or a
relative maximum.
Here’s what it says:
If c is a critical number of a continuous function on an open interval containing c, and if the
function is differentiable on the interval, except possibly at c, then f  c  can be classified as
follows:

f  c  has a relative minimum at c if f   x  changes from negative to positive at c.

f  c  has a relative maximum at c if f   x  changes from positive to negative at c.

f  c  has neither a relative minimum nor a relative maximum at c if f   x  is positive on
both sides of c or is negative on both sides of c.
Example 2
Find the relative extrema of the function f  x  
1
x  sin x on the interval 0, 2  .
2
4
3.5
3
2.5
2
1.5
1
0.5
π
π
2π
3
3
3
0.5
1
1.5
π
4π
5π
3
3
2π
Example 3
Find the relative extrema of f  x    x 2  4 
2/3
8
7
6
5
4
3
2
1
6
4
2
2
1
2
4
6
8
Example 4
Find the relative extrema of f  x  
x4  1
.
x2
9
8
7
6
5
4
3
2
1
4
3
2
1
1
1
2
3
4
CONCAVITY AND THE SECOND DERIVATIVE
Definition of concavity:
If f is a differentiable function on an open interval, then the graph of f is concave upward if f 
is increasing on the interval and concave downward if f  is decreasing on the interval.
According to the definition, to find the open intervals on which the graph of the function f is
concave upward or concave downward, you need to find the intervals on which f  is increasing
or decreasing. To do this, we use the following Test for Concavity:
If f is a function whose second derivative exists on an open interval, then:

If f   0 for all x in the interval, then the graph of f is concave upward on the interval.

If f   0 for all x in the interval, then the graph of f is concave upward on the interval.
Example 5
Determine the open intervals on which the graph of f  x  
6
is concave upward or
x 3
2
downward.
4.5
4
3.5
3
2.5
2
1.5
1
0.5
4
3
2
1
1
0.5
2
3
4
Example 6
Determine the open intervals on which the graph of f  x  
x2  1
is concave upward or
x2  4
downward.
8
6
4
2
15
10
5
5
2
4
6
10
Points of Inflection
The points where the graph of a function changes concavity are called the points of inflection. A
point  c, f  c   is a point of inflection if either f   c   0 or f   x  does not exist at x  c .
To locate possible points of inflection, you can determine the values of x for which f   c   0 or
f   x  does not exist. This is similar to the procedure for locating relative extrema.
Example 7
Find the points of inflection and discuss the concavity of the graph of f  x   x4  4 x3 .
60
50
40
30
20
10
4
3
2
1
1
10
20
30
40
2
3
4
5
6
The Second Derivative Test
In addition to testing for concavity, the second derivative can be used to perform a simple test for
relative maxima and minima. The test is based on the fact that if the graph of a function f is
concave upward on an open interval containing c, and f   c   0 , then f  c  must be a relative
minimum. Similarly, if the graph of a function f is concave downward on an open interval
containing c, and f   c   0 , then f  c  must be a relative maximum.
Here is the Second Derivative Test:
Let f be a function such that f   c   0 and the second derivative exists on an open interval
containing c.

If f   c   0 , then f has a relative minimum at  c, f  c   .

If f   c   0 , then f has a relative maximum at  c, f  c   .
If f   c   0 , the test fails. The function may have a relative maximum, a relative minimum or
neither. In such cases, you can use the First Derivative Test.
Example 8
Find the relative extrema for f  x   3x5  5x3 using the Second Derivative Test.
3
2.5
2
1.5
1
0.5
4
3
2
1
1
0.5
1
1.5
2
2.5
3
2
3
4
Limits at Infinity
Limits at infinity are used to discuss the “end behavior” of a function on an infinite interval. If
the graph of a function approaches a line as x increases without bound, that line is the horizontal
asymptote.
Definition: The line y  L is a horizontal asymptote of the graph of f if
lim f  x   L or lim f  x   L
x 
x
Example 9
Find the horizontal asymptote of each:
2
x2

f  x  5 

f  x 
2x 1
x 1

f  x 
2x  5
3x 2  1

2 x2  5
f  x  2
3x  1

f  x 
2 x3  5
3x 2  1

f  x 
1
x 1

f  x 
2
3x  2
2 x2  1
(This one has two horizontal asymptotes)
Infinite Limits at Infinity
Sometimes the graph of a function does not approach a finite limit as x increases (or decreases)
without bound.

lim f  x    means that the graph rises infinitely as x increases without bound to the
x
right.

lim f  x    means that the graph falls infinitely as x increases without bound to the
x 
right.

lim f  x    means that the graph rises infinitely as x increases without bound to the
x 
left.

lim f  x    means that the graph falls infinitely as x increases without bound to the
x
left.
Example 10
Find each limit and describe what this means about the graph of the function.


lim x3
x 
lim x3
x 
2 x2  4 x
x 
x 1

lim

2x2  4x
x 
x 1
lim
5
30
4
25
3
20
15
2
10
1
5
15
6
4
2
2
4
10
5
5
6
5
1
10
2
15
20
3
25
4
30
10
15
Cusps
Cusps are the points at which a graph makes “abrupt” changes in directions. The derivative does
not exist at these points, although the graph may be continuous at these points. The graphs
below illustrate cusps.
14
14
4
12
12
10
10
8
8
6
6
4
4
2
2
2
2
4
6
8
10
12
2
14
2
2
2
4
4
6
6
4
6
8
If a curve has a cusp at a particular point, the following will be true:

The curve will be continuous at the cusp.

The derivative will not exist at the cusp.

The derivative of the function will approach  at one side of the cusp and will approach
  at the other side of the cusp.
In order to find a cusp, you need to look at points where the first derivative is undefined, and
then evaluate the limit of the derivative as x approaches the cusp from both sides.
Example 11
Determine if the graph of f  x   3  x  1
2/3
 2 has a cusp, and then locate the cusp.
14
12
10
8
6
4
2
4
2
2
2
4
6
4
6
8
10
12
14
Curve Sketching
To sketch a curve, consider this four-part analysis that’ll give you all the information you need.
Test the Function
Find where f  x   0 . This tells you the x-intercepts (or roots). By setting x  0 , we can
determine the y-intercept. Finally, find any horizontal and/or vertical asymptotes.
Test the First Derivative
Find where f   x   0 . This gives you the critical points. We can determine whether the curve
is rising or falling, as well as where the maxima and minima are. It’s also possible to determine
if the curve has any points where it’s nondifferentiable.
Test the Second Derivative
Find where f   x   0 . This shows you the where any points of inflection are. These points are
where the graph of the function changes concavity. Then we can determine where the graph
curves upward and where it curves downward.
Test End Behavior
Look at what the general shape of the graph will be, based on the values of y for very large
values of  x .
Here are things to remember:
1. When f   x   0 , the curve is rising. When f   x   0 , the curve is falling. When
f   x   0 or f   x  does not exist, the curve is at a critical point.
2. When f   x   0 , the curve is “concave up.” When f   x   0 , the curve is “concave
down.” When f   x   0 , the curve is at a point of inflection.
3. The y-coordinates of each critical point are found by plugging the x-value into the
original function.
Example 12
Sketch the graph of f  x   x3 12 x .
Step 1 Find intercepts:
x-intercepts
y-intercept
x3  12 x  0

x( x 2  12)  0
x x  12

y  03  12  0  0

x  12  0
x  0, x   12
There are no asymptotes (polynomial functions do not have asymptotes).
x-intercepts:
 0, 0 ,
y-intercept:
 0, 0

 
12, 0 ,
12, 0

Step 2 Find critical points by taking the derivative of the function.
f   x   3x2 12
Set the derivative equal to zero and solve for x:
3x 2  12  0


3 x2  4  0
3  x  2  x  2   0
x  2
Next, plug x  2 and x  2 into the original equation.
y   2   12  2   8  24  16
3
y   2   12  2   8  24  16
3
Critical points are at  2,  16  and
 2, 16 .
Step 3 Now take the second derivative to find any points of inflection.
f   x   6 x
Set the derivative equal to zero and solve for x:
6x  0
x0
The curve has a point of inflection at  0, 0 .
Now plug the critical values (found in step 2) into the second derivative to determine
whether is a maximum or a minimum. f   2  6  2  12 . This is positive, so the curve
has a minimum at  2,  16  and the curve is concave up at that point.
f   2  6  2  12 . This is negative, so the curve has a maximum at  2, 16 and
the curve is concave down there.
Armed with this information, we can now plot the graph.
Example 13
Sketch the graph of f  x   x4  2 x3  2 x2  1.
Step 1 Find intercepts:
If the equation doesn’t factor easily, it’s best not to bother to find the function’s roots.
y-intercept
y  f  0   0  2 0  2 0  1  1
4
3
2
There are no asymptotes (polynomial functions do not have asymptotes).
x-intercepts:
Not easy to find. Maybe use a calculator or don’t bother.
y-intercept:
 0, 1
Step 2 Find critical points by taking the derivative of the function.
f   x   4 x3  6 x 2  4 x
Set the derivative equal to zero and solve for x:
4 x3  6 x 2  4 x  0


2 x 2 x 2  3x  2  0
2 x  2 x  1 x  2   0
x  0,
Next, plug x  0 , x 
1
, 2
2
1
, and x  2 into the original equation.
2
y  f  0   0  2 0  2 0  1  1
4
3
4
2
3
2
13
1 1
1
1
y  f       2   2  1 
16
2 2
2
2
y  f  2    2   2  2   2  2   1  7
4
3
 1 13 
Critical points are at  0, 1 ,  ,  , and
 2 16 
2
 2,  7  .
Step 3 Now take the second derivative to find any points of inflection.
f   x   12x2  12x  4
Set the derivative equal to zero and solve for x:
12 x 2  12 x  4  0


4 3x 2  3x  1  0
3x 2  3x  1  0
3  21
6
x  0.26,  1.26
x
Now find the y-coordinates for the points above:
y  f  0.26    0.26   2  0.26   2  0.26   1  0.90
4
3
2
y  f  1.26    1.26   2  1.26   2  1.26   1  3.66
4
3
2
The curve has two points of inflection at  0.26, 0.90 and
 1.66,  3.66 .
Now plug the critical values (found in step 2) into the second derivative to determine
whether is a maximum or a minimum.
f   0   12  0   12  0   4  4 .
2
This is negative, so the curve has a maximum at  0, 1 and the curve is concave down at
that point.
2
1
1
1
f     12    12    4  5 . This is positive, so the curve has a minimum at
2
2
2
 1 13 
 ,  and the curve is concave up there.
 2 16 
f   2   12  2   12  2   4  20 . This is positive, so the curve has a minimum at
2
 2,  7 
and the curve is concave up there.
We can now plot the curve.
Example 14
Sketch the graph of f  x   2  x2/ 3 .
Step 1 Find intercepts:
x-intercepts
y-intercept
2  x2/ 3  0
x2/ 3  2
x 
2/3
3/ 2
2
f 0  2  0
3/ 2
2/3
2
x  23/ 2  2 2
There are no asymptotes (there are no places where the function is undefined).
x-intercepts:
2
  2
y-intercept:
 0, 2
2, 0 ,
2, 0

Step 2 Find critical points by taking the derivative of the function.
2
2
f   x    x 1/ 3   3
3
3 x
Set the derivative equal to zero and solve for x:

2
3
3 x
0
We have a problem here. There are no values of x for which the derivative is zero.
However, at x  0 , the derivative is undefined. Therefore, x  0 is a critical number of
the function. We need to see if the graph has a cusp when x  0 . If we look at the limit
as x approaches 0 from both sides, we can determine whether the graph has a cusp.

2 
2
lim   3   
x 0
3
 3 x

 1 
2
 lim  3         
x 0
3
 x 


2 
2
lim   3   
x 0
3
 3 x

 1 
2
 lim  3          
x 0
3
 x 

Therefore, the curve has a cusp at  0, 2 . There are no other critical points.
We can see that when x is negative, the derivative is positive. Therefore, the curve is
rising to the left of zero.
We can also see that when x is positive, the derivative is negative. Therefore, the curve is
falling to the right of zero.
2
Step 3 Now take the second derivative to find any points of inflection. Use f   x    x 1/ 3 .
3
f   x  
2 4 / 3
2
2
2
x
 4/3 

4
9
9x
9  x1/ 3 
9 3x
 
4
Set the derivative equal to zero and solve for x:
2
9
 
3
4
0
x
Again, there are no values of x where this is zero. There are no inflection points. In fact,
the second derivative is positive at all values of x except 0. Therefore, the graph is
everywhere concave up.
We can now plot the curve.
Example 15
Sketch the graph of f  x  
3x
.
x2
Step 1 Find intercepts. A fraction can only equal zero when the numerator is zero. So, to find
the x-intercepts, set the numerator equal to 0 and solve for x.
x-intercepts
3x
0
x2
3x  0
y-intercept
f  0 
x0
x-intercepts:
 0, 0
y-intercept:
 0, 0
30
0  2

0
0
2
Next, look for asymptotes. The denominator is undefined at x  2 , so there is a vertical
asymptote at x  2 . Take the left-hand and right-hand limits of the function at x  2 :
lim
x 2
3x
 
x2
and
lim
x 2
3x
 
x2
This means that the graph rises infinitely to the left of x  2 and falls infinitely to the
right of x  2 .
3x
3x
3
3
 lim x  lim

3.
As x moves infinitely to the right, we have lim
x  x  2
x  x
2 x
2 1 0

1
x x
x
3x
3x
x  lim 3  3  3 .
 lim
As x moves infinitely to the left, we have lim
x  x  2
x  x
2 x
2 1 0

1
x x
x
Therefore, y  3 is a horizontal asymptote.
Step 2 Find critical points by taking the derivative of the function.
f  x 
 x  2  3   3x 1  6
2
2
 x  2
 x  2
Set the derivative equal to zero and solve for x:
6
 x  2
2
0
There are no values of x for which the derivative is zero. Because the numerator is 6 and
the denominator is squared, the derivative will always be positive (the curve is always
rising). You should note that the derivative is undefined at x  2 , but you already know
that there’s an asymptote at x  2 , so you don’t need to examine this point further.
Step 3 Now take the second derivative to find any points of inflection. Use f   x  
 x  2  0   6  2  x  2
2
f   x  
 x  2  


2
2

12  x  2 
 x  2
4

12
 x  2
3
Set the derivative equal to zero and solve for x:
12
 x  2
3
0
Again, there are no values of x where this is zero.
When x  2 , the expression is positive and so the graph is concave up.
When x  2 , the expression is negative and so the graph is concave down.
We can now plot the curve.
6
 x  2
2
.