Summary of First and Second Derivative Tests
The first derivative is used to:

Find stationary points.
Method
Example:
f ( x)  3x  8 x  2
Set the first derivative equal to zero
4
3
f '( x)  12 x 3  24 x 2
0  12 x 3  24 x 2
0  12 x 2 ( x  2)

Determine intervals on which a
function is increasing or
decreasing. You can do this by
using a sign diagram.
Test x-values around the stationary points to
determine the sign or the first derivative.
x  0, 2
There are 2 stationary points: x  0, 2
Test values around x = 0 and x = 2 into the
first derivative. Try x = -1, 1, and 3.
Create a sign diagram.
Here are the signs of the first derivative.
When the first derivative is negative the
function is decreasing. When it is positive, the
function is increasing.


Classify stationary points of a
function (local mins, local maxs,
points of inflection)
The classification can be done
with the first derivative if you use
a sign diagram; otherwise, refer to
the second derivative test for the
classification.
See page 577 of your text for a clearer image of this table.

Since f’(x) is negative before x = 2, we
conclude f(x) is decreasing on x  2 .

Since f’(x) is positive after x = 2, we
conclude f(x) is increasing on x  2 .
Since the sign doesn’t change on either side
of x = 0, it is a horizontal point of inflection
(sometimes also called a stationary point of
inflection.)
Since the sign changes from negative to
positive around x = 2, the graph of f(x)
decreases and then increases. This means
x = 2 is a local minimum.
Of course, if you have your calculator available, you can always verify all of this by looking at the graph of your function. 
IBSL2
Peabody 2009-10
The second derivative is used to:

Find all points of inflection.
Method
Example:
f ( x)  3x  8 x  2
Find all the points of inflection by setting the
second derivative equal to zero.
4
3
f '( x)  12 x 3  24 x 2
f "( x)  36 x 2  48 x
0  36 x 2  48 x
0  12 x(3 x  4)
x  0,

Classify points of inflection (either
horizontal or stationary points of
inflection or non-horizontal, nonstationary points of inflection)
Note that in either case, a point of
inflection is a point at which the
concavity of the graph changes.

Determine the intervals on which a
function is concave up or down.
You can do this through use of a
sign diagram testing signs of the
second derivative on either side of
the points of inflection.
4
3
If both first and second derivative are zero,
then the point is a horizontal point of
inflection (or stationary point of inflection).
Since f '(0)  0 and f "(0)  0 , x = 0 is a horizontal
point of inflection.
If only the second derivative is zero, then the
point is a non-horizontal (or non-stationary)
point of inflection.
Since f '( 43 )  0 while f "( 43 )  0 , x  43 is a nonhorizontal (or non-stationary) point of inflection
Test x-values around the points of inflection
to determine the sign or the second
derivative.
Test values around x = 0 and x  43 into the second
derivative. Try x = -1, 1, and 2.
Create a sign diagram.
Here are the signs of the second derivative.

+
0
When the second derivative is positive, the
function is concave up. When it is negative
the function is concave down.
IBSL2
+
4
3

Since f’(x) is positive before x = 0 and after
x  43 the graph is concave up on the intervals
x  0, and x  43

Since f’(x) is negative between x = 0 and x  43 ,
the graph is concave down on the interval
0  x  43 .
Peabody 2009-10

Classify stationary points of a
function (local mins, local maxs,
points of inflection)
To use the second derivative to classify
stationary points, simply test the stationary
point into the second derivative.
If the result is positive, then the graph is
concave up. Hence, the stationary point is a
local minimum.
If the result is negative, then the graph is
concave down. Hence the stationary point is
a local maximum.
f ( x)  3x 4  8 x3  2
f '( x)  12 x 3  24 x 2
f "( x)  36 x 2  48 x
 12 x(3 x  4)
From our earlier work, we found two stationary
points: one at x = 0 and the other at x = 2.
Testing these into the second derivative, we get:
For x = 0:
If the result is zero, then the point is a
horizontal point of inflection.
f "( x)  12 x(3x  4)
f "(0)  0(4)
So x = 0 is a point of inflection.
0
Note that we needed a sign chart with the first
derivative to come to this conclusion earlier.
f "( x)  12 x(3 x  4)
For x = 2:
f "(2)  24(6  4)
 positive
Since the result is positive, the graph is concave up
at x = 2. This means that x = 2 is a local minimum.
Again, the same conclusion can be reached by
using a sign chart with the first derivative.
This is the real math-y stuff you need for the quiz. You also need to understand how all of it relates to physical situations involving
displacement, velocity and acceleration, etc. The charts on page 567 of your textbook seem particularly useful for this. If you are looking for
extra problems for review, you should be able to now complete the following from your text:
Review Set 22A: Q1, Q2, Q3, Q4, Q6, Q8, Q11
Review Set 22B: Q2, Q3, Q4, Q6
IBSL2
Peabody 2009-10