Relative Extrema: Graphing
Polynomials
Objective: We will locate relative
maximum and minimum values and use
all of our knowledge to graph
polynomials.
Definition 5.2.1
• A function f is said to have a relative maximum at x0
if there is an open interval containing x0 on which
f(x0) > f(x) for all x in the interval. Similarly, f is said
to have a relative minimum at x0 if there is an open
interval containing x0 on which f(x0) is the smallest
value, that is, f(x0) < f(x) for all x in the interval
Definition 5.2.1
• A function f is said to have a relative maximum at x0
if there is an open interval containing x0 on which
f(x0) > f(x) for all x in the interval. Similarly, f is said
to have a relative minimum at x0 if there is an open
interval containing x0 on which f(x0) is the smallest
value, that is, f(x0) < f(x) for all x in the interval. If f
has either a relative maximum or a relative minimum
at x0 , then f is said to have a relative extremum at x0.
Example 1
• f(x) = x2 has a relative min at x = 0 but no max.
Example 1
• f(x) = x2 has a relative min at x = 0 but no max.
• f(x) = x3 has no relative extrema.
Example 1
• f(x) = x2 has a relative min at x = 0 but no max.
• f(x) = x3 has no relative extrema.
• f(x) = x3 – 3x + 3 has a relative max at x = -1 and a
relative min at x = 1.
Example 1
• f(x) = x2 has a relative min at x = 0 but no max.
• f(x) = x3 has no relative extrema.
• f(x) = x3 – 3x + 3 has a relative max at x = -1 and a
relative min at x = 1.
• f(x) = x2/3 has a relative min at x = 0 but no max.
Critical Points/Stationary Points
• The relative extrema for the four functions we just
saw occur at points where the graphs of the function
have a horizontal tangent or at points of nondifferentiability.
• We define a critical point for a function f to be a point
in the domain of f at which there is either a
horizontal tangent or a point of non-diff.
• More specifically, we call the points where there is a
horizontal tangent a stationary point.
Theorem 5.2.2
• Suppose that f is a function defined on an open
interval containing the point x0. If f has a relative
extremum at x = x0, then x = x0 is a critical point of f;
that is, either f /(x0) = 0 or f is nondifferentiable at x0.
Example 2
• Find all critical points of f(x) = x3 – 3x + 1. State if it
is a stationary point.
Example 2
• Find all critical points of f(x) = x3 – 3x + 1. State if it
is a stationary point.
• f /(x) = 3x2 – 3
= 3(x + 1)(x – 1)
• The critical points occur at
• x = 1 and x = -1 and they are
• both stationary points.
Example 3
• Find all critical points of f(x) = 3x5/3 – 15x2/3. State if
it is a stationary point.
Example 3
• Find all critical points of f(x) = 3x5/3 – 15x2/3. State if
it is a stationary point.
• f /(x) = 5x2/3- 10x-1/3
= 5x-1/3(x – 2)
• x = 0 is a point of non-diff,
• so it is a critical point,
• and x = 2 is a zero of the
• first derivative, so it is
• a stationary point.
First Derivative Test
• We have established that a relative max/min must
occur at a critical point. But every critical point is not
necessarily a relative max/min. Lets look at some
different graphs.
First Derivative Test
Critical point
Stationary pt.
Relative max
Pos to neg
Critical point
No stat. pt.
Relative max
Pos to neg
Critical point
Stationary pt.
Relative min
Neg to Pos
First Derivative Test
Critical point
Stationary pt.
Inflection pt.
No max/min
Critical point
Stationary pt.
Inflection pt.
No max/min
Critical point
No stat. pt.
Inflection pt.
No max/min
First Derivative Test
•
A function has a relative extremum at those critical
points where f / changes sign.
• Theorem 5.2.3 (First Derivative Test)
a) If f /(x) has a critical point at x and the first
derivative changes sign from positive to negative at
x, there is a relative max at x.
b) If f /(x) has a critical point at x and the first
derivative changes sign from negative to positive at
x, there is a relative min at x.
Example 4
• We looked at f(x) = 3x5/3 – 15x2/3 in example 3. Lets
do a sign analysis to see if the critical points are
relative extremum.
• f /(x) = 5x2/3- 10x-1/3
= 5x-1/3(x – 2)
______+___|__-___|__+_____
0
2
r max r min
c.p.
s.p.
Second Derivative Test
• We have another way to determine if a critical point
is a relative max or min. We know that the second
derivative tells us about concavity. If a function is
concave up at a critical point, that would be a
minimum. If a function is concave down at a critical
point, that would be a maximum.
Second Derivative Test
• Theorem 5.2.4 (Second Derivative Test)
• Suppose that f is twice differentiable at the point x.
a) If f /(x) = 0 and f //(x) > 0, then f has a relative
minimum at x.
b) If f /(x) = 0 and f //(x) < 0, then f has a relative
maximum at x.
c) If f /(x) = 0 and f //(x) = 0, then the test is
inconclusive. There could be a max, a min, or
neither at x.
Example 5
• Find the relative extrema of f(x) = 3x5 – 5x3.
Example 5
• Find the relative extrema of f(x) = 3x5 – 5x3.
• f /(x) = 15x4 – 15x2
= 15x2(x + 1)(x – 1)
• Using the first derivative test:
____+_____|___-___|__-___|__+____
-1
0
1
r max
r min
s.p.
s.p.
s.p.
Example 5
• Find the relative extrema of f(x) = 3x5 – 5x3.
• f /(x) = 15x4 – 15x2
= 15x2(x + 1)(x – 1)
• Using the second derivative test:
• f //(x) = 60x3 – 30x
= 30x(2x2 – 1)
Example 5
• Find the relative extrema of f(x) = 3x5 – 5x3.
• f /(x) = 15x4 – 15x2
= 15x2(x + 1)(x – 1)
• Using the second derivative test:
• f //(x) = 60x3 – 30x
= 30x(2x2 – 1)
f //(-1) = negative, concave down, relative max
f //(0) = 0; inconclusive
f //(1) = positive, concave up, relative min
Multiplicity
• Roots of different multiplicity look different on a
graph. The most common that we will deal with are
single, double, and triple roots. In general, odd roots
will have a sign change and cross the x-axis and even
roots will not have a sign change and will not cross
the x-axis.
Example 7
• We will now use everything that we have learned to
graph a polynomial function without a calculator.
• Graph y = x3 – 3x + 2 using the first and second
derivative.
Example 7
• We will now use everything that we have learned to
graph a polynomial function without a calculator.
• Graph y = x3 – 3x + 2 using the first and second
derivative.
• f /(x) = 3x2 – 3 = 3(x + 1)(x – 1)
Inc
Dec
Inc
____+____|____-____|____+____
-1
1
max
min
sp
sp
Example 7
• We will now use everything that we have learned to
graph a polynomial function without a calculator.
• Graph y = x3 – 3x + 2 using the first and second
derivative.
• f //(x) = 6x
C down
C up
____-____|____+____
0
IP
Example 7
• If possible, find the x and y intercepts. In this
problem, the y-intercept is easy (0, 2).
• The x-intercepts are more difficult. We can find them
using the rational root theorem and synthetic
division. The choices for rational roots are +1, + 2.
1| 1 0 -3 2
1 1 -2 |0
(x+2)(x-1); roots x = 1
• Find the max/min values.
x = -2
f(-1) = 4, f(1) = 0
Example 7
Inc
Dec
Inc
____+____|____-____|____+____
-1
1
max
min
C down
C up
____-____|____+____
0
y-intercept is (0, 2)
Homework
• Pages 287-288
• 1-37 odd