Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 42 pps

EXAMPLE 11.6 Graph f x = x3− 2x2− 8x, identifying the x-coordinates, all zeros, local extrema, and inflection points.. By drawing a number line showing these points where fx = 0 and indi

11.3 POLYNOMIAL FUNCTIONS AND THEIR GRAPHS

Graphing Polynomials

We’ll begin with a couple of examples and then arrive at a general strategy for understanding the graphs of polynomials

EXAMPLE 11.6 Graph f (x) = x3− 2x2− 8x, identifying the x-coordinates, all zeros, local extrema, and

inflection points

SOLUTION Zeros. This is the function from Example 11.2 In factored form f (x) = x(x − 4)(x + 2);

it has zeros at x = −2, x = 0, and x = 4 The graph of f (x) does not cross the x-axis anywhere between these zeros f is continuous, so its sign can’t change between the zeros Therefore, between x = −2 and x = 0, for example, f (x) must be always positive or always negative By determining the sign of f (x) for just one test value of x on the interval (−2, 0),

we can determine the sign of f (x) on the entire interval (−2, 0) We draw a number line showing the zeros of f (x) and the sign of f (x) between those zeros

Sign Computations—suggested method:Work with f in factored form:

f (x) = x(x − 4)(x + 2) We’ll start with x very large and then look at the effect on the sign

of the factors as x decreases and passes by each zero

(–) –2 (+) 0 (–) 4 (+)

Sign of f

For x very large, all factors are positive: (+)(+)(+) ⇒ (+)

As x drops below 4, the factor (x − 4) changes sign: (+)(−)(+) ⇒ (−)

As x drops below 0, the factor x changes sign: (−)(−)(+) ⇒ (+)

As x drops below −2, the factor (x + 2) changes sign: (−)(−)(−) ⇒ (−)

A spot-check can be used to confirm this

4 –2

y

x

Figure 11.10

Local Extrema. We need to look at the first derivative to determine critical points: f(x) = 3x2− 4x − 8 We use the quadratic formula to determine where f(x) = 0 and obtain

x =2+2

√ 7

3 ≈ 2.43, and x =2−2

√ 7

3 ≈ −1.10 By drawing a number line showing these points where f(x) = 0 and indicating the sign of f(x)between them, we can determine whether these stationary points are local minima, local maxima, or neither

graph of f(x) sign of f ′(x) 2 – 2√7

3

2 + 2√7 3

Computations:

If x is large enough in magnitude, x2dominates and fis positive

For x = 2/3 (between the zeros) fis negative

So, f (x) has a local maximum at x =2−23√7and a local minimum at x = 2+23√7

We know that a cubic on a nonrestricted domain will have neither a global maximum value nor a global minimum value; when the domain of a cubic is (−∞, ∞), the range is also (−∞, ∞)

Inflection Points. Take the second derivative: f(x) = 6x − 4 f(x) = 0 when x = 2/3

We draw another number line to show the sign of f(x)and the corresponding concavity

of f (x)

concave down concave up (–) 2 (+) 3

concavity of f(x) sign of f ″(x)

All this information taken together enables us to draw an accurate graph of f (x)

y

x

5 3

4 –2 2–2√7

3

2+2√7 3 2

f (x)

Figure 11.11

EXAMPLE 11.7 Sketch a rough graph of f (x) = −(x + 3)(x − 2)2, labeling all zeros and showing where

f (x)is positive and where f (x) is negative See how much information can be gotten without using derivatives

SOLUTION The cubic f (x) has only two zeros: a zero at x = −3 and a double zero at x = 2 As in

Example 11.6, let’s draw a number line showing the zeros of f (x) and the sign of f (x) in between these zeros

2 –3

y

x

Figure 11.12

Notice that here f (x) does not change sign when it passes through the zero at x = 2; it is negative on both sides of this zero This is because the (x − 2)2term in f (x) doesn’t change sign when x passes through x = 2

y

x

2 –3

f (x)

Figure 11.13

f has a local maximum at x = 2 To find the local minimum, we would need to calculate

f Do this as an exercise

Can you figure out in general what effect multiple (double, triple, or higher) roots have

on the graph of f (x)? (Hint: Consider separately the cases where the factor (x − a) is raised

to even and odd powers and analyze the sign.)

Strategy for Graphing a Polynomial Function f

1 Bird’s-eye view

Look at the degree of the polynomial and the sign of the leading coefficient to anticipate what is ahead From this information, determine whether the polynomial will have a global maximum or a global minimum or neither

2 Positive/negative

x-intercepts are simple to find if we can factor the polynomial into linear and quadratic factors In theory this is always possible, but in practice it may be very difficult Polynomials are continuous functions, so if P (x) changes sign between x = a and

x = b there must be at least one x-intercept between x = a and x = b If you find the x-intercepts, plot them If you are certain you’ve identified all the zeros of P (x), then check the sign of P (x) in each interval.5

Note:If the x-intercepts are symmetrically placed, check for even or odd symmetry

If x is only raised to even powers, the graph will be symmetric about the y-axis If x

is only raised to odd powers and there is no constant term, then the graph will be symmetric about the origin

y-intercept: P (0)

3 Increasing/decreasing

Look at the sign of P(x) On a number line plot the points where P(x) = 0 and then check the sign in every interval

P>0 ⇒ the graph of P is increasing

P<0 ⇒ the graph of P is decreasing

5 If the polynomial is in factored form, it’s easy to check the sign of the polynomial at, say, x = 3, by checking the sign of each of the factors at x = 3 and then “multiplying the signs” together.

Since Pis continuous, P= 0 at any turning point (local maximum or minimum) of the graph

Note:If P(3) = 0 we cannot conclude that the graph must have a local maximum or minimum at x = 3 We can only be sure that the graph has a horizontal tangent line at

x = 3 P will have a local extrema at x = 3 only if Pchanges signat x = 3

4 Concavity—fine tuning Look at the sign of P(x) On a number line plot the points where P(x) = 0 and check the sign in every interval

P>0 ⇒ the graph of P is concave up

P<0 ⇒ the graph of P is concave down

Note: P will have a point of inflection at x = c only if Pchanges signat x = c

5.Look back at what you’ve done Consider the degree of the polynomial and make sure that your graph makes sense

A graphing calculator or computer can be useful for graphing polynomials For in-stance, if zeros are hard to find a machine is useful in approximating them It is up to you, however, to determine an appropriate viewing window

EXAMPLE 11.8 Graph6P (x) = 100x4− 900x2

SOLUTION 1 P (x)is a fourth degree polynomial with a positive leading coefficient, so from a

bird’s-eye view the graph should look something like

.

At the extremities the graph reaches upward The graph must have either one or three turning points.7

2 P (x) = 100x4− 900x2= 100x2(x2− 9) = 100x2(x + 3)(x − 3) The x-intercepts are therefore at x = 0, x = 3, and x = −3 (Plot these zeros.) The x-intercepts are symmetrically located, so we check for symmetry All x’s are raised to even powers, so P (−x) = P (x) This means the graph has even symmetry; it

is symmetric about the y-axis

3 P (x) = 100(x4− 9x2), so P(x) = 100(4x3− 18x) = 200x(2x2− 9)

6 Try to graph this on your graphing calculator You’ll probably find that it takes you some time to find a suitable viewing window.

7This follows from the previous statement The derivative of a fourth degree polynomial is a cubic, which can have one, two,

or three zeros, but the cubic will change sign either once, or three times.

or

y

x

y

x y

x

3 –3

p

x

P(x) = 0 at x = 0 and where 2x2− 9 = 0, i.e., x2=92, or x = ±√3

2

graph of P sign of P′ (–) –3 (+) 0 (–) (+)

There are local minimum points at x =√3

2 and at x = −√3

2 and a local maximum at

x = 0

P (x) = 100x2(x2− 9), so

P (0) = 0 P

 3

√ 2



= 100 9

2

  9

2 − 9



= 450



−9 2



= −2025

Likewise,

P



−√3 2



= −2025

4 P(x) = 100(12x2− 18) = 200(6x2− 9)

P(x) = 0 provided that

6x2= 9

x2=9 6

x2=3 2

x = ± 3

2.

graph of P sign of P ″ (+)– 3 (–) (+)

2

3 2

concave up

concave up

concave down

First, using the information we have gathered, try to draw the graph yourself Now use a graphing calculator or computer to check your answer Notice that you have to be careful about your choice of range and domain, otherwise the machine will be of little help

√2

3

√2

– 3 2

2

√

P

x

Figure 11.14
EXERCISE 11.8 Consider the polynomial f (x) = 100x4− 900x2− 2

(a) How many zeros does f (x) have? (Use the results of Example 11.8 to aid you.) (b) Approximately where are the zeros? (Give your answers to within 0.01.) (c) Show that 3 is not a zero of f (x) Is the zero in the vicinity of 3 larger than 3 or smaller than 3? Explain your reasoning

Answer

There are two zeros near x = ±3 In actuality, one is between 3.00031 and 3.00052, and the other is between −3.00031 and −3.00052

EXERCISE 11.9 Graph y = 100x4− 900x2+ 800 (Hint: Factor this first into two quadratics, then factor

the quadratics themselves.)

Finding a Polynomial Function to Fit a Polynomial Graph

EXAMPLE 11.9 Find a function to fit the cubic function f (x) shown

(2, –5)

x

y

f (x)

Figure 11.15

SOLUTION We see that f (x) has zeros at x = −2, x = 0, and x = 1, so f (x) must have factors (x + 2),

x, and (x − 1) Since it is a cubic, we can write

f (x) = k(x + 2)(x)(x − 1)

Multiplying by the constant k does not change the location of any of the x-intercepts (since it rescales vertically), but it does allow us to ensure that f (x) passes through the point (2, −5)

as it does in the graph Knowing that x = 2, y = −5 must satisfy the equation gives us

−5 = k(2 + 2)(2)(2 − 1)

−5 = 8k

k = −5

8. The equation must be f (x) = −58(x + 2)(x)(x − 1) You can verify this on your graphing calculator

EXAMPLE 11.10 Find a possible equation for the polynomial function g(x) shown

–1

–1.5 –2

1 2 3

–10

g (x)

x y

Figure 11.16

SOLUTION g(x)has three zeros, at x = −1.5, x = 1, and x = 3 Notice that g(x) does not change sign

when it passes through x = −1; instead it “bounces off” the x-axis at this point Therefore

x = −1 must be a root of even multiplicity of g(x) = 0, as in Example 11.7 above We write g(x) = k(x + 1.5)2(x − 1)(x − 3) A bird’s-eye perspective confirms this is reasonable g(x)could be a fourth degree polynomial with a negative leading coefficient

The graph passes through the point (0, −10); we’ll use that to find k

g(x) = k



x +3 2

2

(x − 1)(x − 3)

−10 = k 3

2

2

(−1)(−3)

−10 = k ·274

−40

27= k This gives us

g(x) = −40

27



x +3 2

2

(x − 1)(x − 3)

You can verify that this is a reasonable fit for the graph by using a graphing calculator or computer

EXERCISE 11.10 In Example 11.10 we arrived at the function

g(x) = −40

27



x +3 2

2

(x − 1)(x − 3)

Use your graphing calculator to observe the effect of the following variations

h(x) = −40

27



x +3 2

2

(x − 1)3(x − 3)

j (x) = −160

243



x + 3 2

4

(x − 1)(x − 3)

Explain your observations by taking a numerical perspective

EXERCISE 11.11 Find a possible equation for the cubic function h(x) shown

–2

y

x

h (x)

Figure 11.17

Basic Strategy for Fitting an Equation to a Polynomial Graph

Secure the x-intercepts: If they are at x = a, x = b, and x = c, start with P (x) = k(x − a)(x − b)(x − c)

See whether or not the sign of y changes on either side of the x-intercept

The factor (x − a) changes sign as x increases from values less than a to those greater than a

If the sign changes around x = a, then the factor (x − a) must be raised to an odd power

If the sign of y does not change around x = a, then the factor (x − a) must be raised

to some even power.8

If you know both the x- and y-coordinates of some point that is not a zero, use this information to find k If this information is unavailable, check the sign somewhere to determine the sign of k

8 Generally it’s wise to begin with the lowest power that has the appropriate effect (Doing Exercise 11.10 will help you with this.) Below are figures illustrating the effect of (x − a) versus (x − a) 3

in the neighborhood of x = a Increasing the power of (x − a) will flatten the graph in the immediate vicinity of x = a and make it steeper far away from x = a.

(x – a)3

Take a bird’s-eye view to make sure your answer is reasonable Look at the degree and the leading coefficient

REMARK This strategy is good for finding equations to fit graphs where nothing too exciting

or extravagant is happening between the zeros It won’t help substantially in any of the cases sketched below

y

x

y

x

y

x

f (x)

Figure 11.18

P R O B L E M S F O R S E C T I O N 1 1 3

1 Suppose you are given a polynomial expression in both factored and nonfactored form When might you prefer one form over the other?

2 Suppose P (x) is a polynomial whose derivative is P(x) = x2(x + 2)3

(a) What degree is P (x)?

(b) What are the critical points of P (x)?

(c) Does P (x) have an absolute minimum value? If so, where is it attained? Is it possible to find out what this minimum value is, if it exists? If yes, explain how;

if no, explain why not

3 Let p(x) be a polynomial of degree n What is the maximum number of points of inflection possible for the graph of p(x)?

4 A company is producing a single product P (x), the profit function, gives profit as a function of x, the number of hundreds of items produced Suppose P (0) = −200 and

P(x) = x2(x − 1)3

Sketch the graph of P Argue, using the sign of P(x), that the graph of P intersects the positive x-axis exactly once, i.e., for x > 0, that there is one and only one break-even point and that, if production levels are high enough, the profit will remain positive and increase with increasing x The following questions will help guide you

(a) First, draw a rough sketch of the graph of P(x) (You need not determine precisely the position of the local minimum of P; in other words, you need not take the second derivative—just use what you know about the intercepts and sign of P(x).) (b) Draw a number line and on it record the sign of P Above it indicate where P is increasing and decreasing

(c) Now, using the information that P (0) = −200 along with the information from part (b), make a rough sketch of P You need not determine the positive x-intercept, just convincingly assert that it exists

5 Suppose P (x) with domain (−∞, ∞) is a polynomial of degree 4 whose leading coefficient is −3 For each statement given below, determine whether the statement is

necessarily true, or possibly true, possibly false, or definitely false

Think carefully This is a problem concerning both logic and polynomials

(a) limx→∞P (x) = ∞ (b) limx→−∞P (x) = −∞

(c) P (x) has four zeros

(d) P (x) has at least one turning point

(e) P (x) has exactly two turning points

(f ) P (x) has four critical points

(g) P (x) has an absolute maximum value

(h) P (x) has an absolute minimum value

6 Suppose P (x) is a polynomial of degree 7 whose leading coefficient is 2 For each statement given below, determine whether the statement is

necessarily true, or possibly true, possibly false, or definitely false

If you decide the statement is not necessarily true, explain your reasoning!

(a) P (x) has at least one zero and at most seven zeros

(b) P(x)has no zeros

(c) P (x) has at least one point of inflection

(d) P (x) has five points of inflection

7 Graph f (x) = x44 + x3− 5x2 Indicate the x-coordinates of all local extrema and all points of inflection What is the absolute minimum value of f ? The absolute maximum value?

8 Find equations that could correspond to the graphs of the polynomials drawn below

y

x

(a)

1 2 –2

3

y

x

(b)

1 2 –2

–3

y

x

(c)

–1 –2

(1, 2)