Critical and
Inflection Points
1 Finding and Classifying Critical Points
A critical point is a point on the
graph where the tangent slope
is horizontal, (0) or vertical,
(∞). or not defined like the
minimum of y = |x|.
4
3
2
1
-4
-2
0
2 x
4
Some people call critical
points stationary points or extrema but we will always call
them critical points. First
what’s so special about critical
points? Why are they critical? They are critical because
they could be maximum or
minimum points.
If the tangent slope at a
2
point is horizontal, that is,
y 0 = 0, then the curve has one
of four shapes near that critical
point.
You can see that the first
two situations are a maximum
and minimum respectively.
The bottom two situations are
3
neither, they are just places
on the graph with a horizontal
tangent slope.
If the tangent slope at the
critical point is vertical, that
is, y0 = ∞, then there are
four possible shapes near the
critical point.
The top two are called
4
cusps. The first one has a
critical point that is a maximum and has a vertical tangent
slope. The second is a minimum and also has a vertical
tangent slope. The bottom two
are just critical points where
the tangent slope is vertical.
They are neither maximum nor
minimum.
Ok, let’s do the first part of
this activity, find the critical
points. All we have to do
is take the derivative of the
function in question and see
5
what x0s make the tangent
slope 0 or ∞. That is, set the
derivative equal to zero, or
see where the derivative is
undefined. By undefined we
usually mean the denominator
of a fraction is zero. Let’s do
some examples to flesh this
out.
Example
Q?
Find the critical
points of the function y =
2x3 − 9x2 + 12x + 6.
A.
Take the derivative
6
of the function to get
y 0 = 6x2 − 18x + 12
Now, the derivative or tangent
slope function is a polynomial
so y0 is never going to be ∞.
Set y0 = 0 and solve.
0 = 6x2 − 18x + 12
= 6(x − 1)(x − 2)
Therefore at the points where
x = 1 and x = 2 the tangent
slope is horizontal. This gives
us the x-values of the two
critical points. What are the
y-values? We get the y-values
by plugging the x-values into
7
the original equation, y =
2x3 − 9x2 + 12x + 6. Thus the two
critical points are (1, 11) and
(2, 10).
Now notice the graph of the
function from Example and
spot the critical points.
14
12
10
8
6
0
0.5
1
1.5
x
2
2.5
3
Example
Q?
Find the critical
8
points of the function
A.
1
2 3
y = x 2 − 2x 2
3
Take the derivative
1
1
0
−
2
y =x −x 2
and then factor out the “x to
the lowest power”, in our case,
1
−
x 2.
i
h
1
1
0
−
y = x 2 x2 − 1
=
h 1
i
x2 − 1
x
1
2
We can see that if x = 0, then
the denominator of the above
fraction would be zero, but
x = 0 is a defined point on the
9
graph. Plug x = 0 into the
original equation and see that
the graph goes through the
origin. We conclude that at the
critical point where x = 0 we
have a vertical tangent slope.
To check for horizontal tangent
slopes we set y0 = 0 and solve.
Note that for a fraction to
be zero, the top part must be
10
zero.
y0 =
0=
i
h 1
x2 − 1
1
2
h 1x i
x2 − 1
1
2
x
0 = x −1
1
2
1=x
1
2
x=1
Therefore (1, − 43 ) is a critical
point where the tangent slope
is horizontal.
When we look at the graph
of the function from Example
we can see the critical points
but we also notice that there
11
are no points where the xvalues are negative. Why?
1
0.5
-1
0
1
x2
3
4
-0.5
-1
Because the domain of the
function is [0, ∞). Look back
to the beginning of the term
to refresh your recollection of
range and domains.
Ok, now we can find critical
points. The next step is to
classify them. We need to
know whether a critical point
12
is a
• local maximum
• local minimum
• or neither
To do this we will draw a
slope line. This is an important
illustrative tool. After the
slope line is drawn we will
use the First Derivative Test
to classify the critical points.
There is a Second Derivative
Test that we will learn later
Example
Q?
Classify the critical
13
points of the function y =
2x3 − 9x2 + 12x + 6
1.1 First Derivative
Test
After we draw the slope line
we can use these patterns to
classify the critical points
• / - \ indicates a maximum
• \ - / indicates a minimum
14
/ indicates a cusp that’s a
minimum
• \|
•
/ | \ indicates a cusp that’s a
maximum
•
/ - / indicates a critical point
that is neither a maximum
nor a minimum
- \ indicates a critical point
that is neither a maximum
15
• \
nor a minimum
•
/ | / indicates a critical point
that is neither a maximum
nor a minimum
indicates a critical point
that is neither a maximum
nor a minimum
• \|\
16
Steps for drawing a Slope
Line
1. Find all the critical points of
the function.
2. Draw a number line with the
critical points on it and their
slopes represented above, either as a horizontal or vertical
line.
3. Pick test points ( x-values )
in between the critical points
and plug them into the first
derivative to tell if the tangent
slope is positive or negative in
that interval.
17
4. Add those slopes to the number line to represent the tangent slopes in the intervals
5. Use the First derivative test to
classify the critical points.
This concept of a slope line
can answer many questions
concerning a particular function. We have already used it
to classify critical point with
the First Derivative Test. We
can also use the slope line
to find intervals of increase
or decrease, that is, intervals
18
where the graph is increasing
or decreasing respectively.
Example
Q?
Find the intervals
of increase and decrease of the
function y = 2x3 − 9x2 + 12x + 6
from Example .
Another benefit of the slope
line is that we get a rough idea
of the shape of the graph. The
slope line shown in Example
implies the curve increases
until x = 1 where it reaches
a peak and drops down until
x = 2 where it bottoms out
19
and increases continually after
x = 2. This implication is
confirmed by the actual graph
shown after Example .
Local versus Global
We use the terms local and
global to describe maximum
or minimum points. All maximum or minimum points are
local maximums or minimums
respectively. They are the
highest (or lowest) spot on the
graph in their immediate vicinity. The highest point on the
20
whole graph is the global maximum. Likewise, the lowest
point on the whole graph is the
global minimum.
2 Concavity and
Points of Inflection
The first derivative is the tangent slope formula, the second
derivative is the concavity formula. Concavity is the bend
in the curve. Positive concavity or concave up occurs when
y 00 > 0 and negative concav21
ity or concave down occurs
when y00 < 0. There can be
any combination of slope and
concavity.
• Positive concavity with positive slope, y00 > 0, y0 > 0
•
Positive concavity with negative slope, y00 > 0, y0 < 0
•
Negative concavity with
positive slope, y00 < 0, y0 > 0
22
•
Negative concavity with
negative slope, y00 < 0, y0 < 0
A good way to remember
the difference between positive concavity and negative
concavity is this cheesy child’s
drawing.
Anyway, just like when we
drew the slope line, we draw
a concavity line. The steps
23
to make it are similar to the
slope line. Get the second
derivative of the function and
find the x-values that make y00
zero or undefined and mark
those points on your concavity
line. These points are called
possible points of inflection.
Use test points on either sides
of these possible points of
inflection to determine the
concavity of the intervals.
Wait you say. What’s an
inflection point?
Definition 1
An inflection point is
24
a point on the graph where the
concavity changes from concave
up to concave down or from concave down to concave up.
So a possible point of inflection still has to prove itself
to be a point of inflection by
having different concavities on
either side of it.
Example
Q?
Find the intervals
of positive concavity and the
intervals of negative concavity
25
of the function
x5 x4
− + 7x
y=
20 12
and any inflection points.
A.
We can see the concavity in
the graph of the function from
Example .
100
50
-4
-2
0
2 x
4
-50
-100
Oh, I promised we would
learn the Second Derivative Test, eh. Here it is, as
26
promised.
A point on a curve with a
horizontal tangent slope and
negative concavity must be a
maximum.
Likewise a point with a
horizontal tangent slope and
positive concavity must be a
minimum.
This thinking forms the
basis for the second derivative
27
test. First find the critical
points, then test them with the:
2.1 Second Derivative
Test
To test a critical point x =
a, y 0(a) = 0 to classify it as a
maximum or a minimum:
• If y 00(a) < 0 then x = a is a
maximum point
• If y 00(a) > 0 then x = a is a
minimum point
That’s it. This test is used
frequently in economics. So
28
now we have two ways to
classify critical points. My
personal favorite is the First
Derivative Test. It’s much
better. You only have to take
one derivative. Notice the
Second Derivative Test is only
good for critical points where
the slope is horizontal and
there is no conclusion to be
made when y00(a) = 0. Then
what? You have to do the
First Derivative Test anyway.
Also, you can get so much
more information from the
29
slope line than you can with
the Second Derivative Test.
I’ll show you an example of
the inferior Second Derivative
Test just for completeness.
Example
Q?
Find and classify
the critical points of the curve
y = x4 − 2x2 + 3.
Homework
Section 2.10
#10, 13, 14 - 18, 21, 22
Submit 10, 16, 22
Section 3.4
#25 - 29
Submit 26
30
Section 3.5
#49, 50
Section 3.6
#3 - 12
Submit 4, 6, 10
31