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 non-differentiability.

• 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 Critical point Critical point Stationary pt. No stat. pt. Stationary pt. Relative max Relative max Relative min Pos to neg Pos to neg Neg to Pos

First Derivative Test

Critical point Critical point Critical point Stationary pt. Stationary pt. No stat. pt. Inflection pt. Inflection pt. Inflection pt. No max/min No max/min 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 7Inc Dec Inc ____+____|____-____|____+____ -1 1 max min

C down C up ____-____|____+____ 0

y-intercept is (0, 2)

Homework

• Pages 287-288• 1-37 odd