M17 Zeroes and Inverse Function.pdf

Zeros of Polynomials

and Inverse Functions

At the end of this lecture, a student must be able to:

Relate the zeros of a polynomial with solving equations and factoring

Use the rational zero test to list down possible zeros of a polynomial

Determine the inverse of a function (if it exists) Recognize one-to-one functions

RECAP:

Given polynomial function p(x) with real coefficients of degreen:

• ri is a zero of p(x) ⇔ p(ri) = 0 ⇔x−ri is a factor of p(x)

• p(x) has exactly n zeros (counting multiplicities): p(x) = an(x−r1)m1(x−r2)m2...(x−rk)mk

• Ifx−(a+bi) is a factor, then so is x−(a−bi). • p(x) is a product of linear and quadratic factors with

real coefficients.

Real Zeros of a Polynomial

Let

p

(

x

) be a polynomial function with real

coefficients with domain

R

.

The following are equivalent:

1.

finding all real zeros of

p

(

x

)

2.

obtaining all the real solutions of the

equation

p

(

x

) = 0

3.

finding the

x

-intercepts of the graph of

p

(

x

)

Complex Zeros of a Polynomial

Let

p

(

x

) be a polynomial function with

complex

coefficients

with domain

C

.

The following are equivalent:

1.

finding all complex zeros of

p

(

x

)

2.

factoring

p

(

x

) into linear factors with

complex coefficients

Rational Zero Test

Given apolynomial with only integer coefficients (with an6= 0)

p(x) =anxn+an−1xn−1+· · ·+a1x+a0,

if p

q is a zero ofP(x)and pand q are integers such that p q

is in lowest terms, then p is a factor ofa0 and q is a factor

of an. Note:

1. Ifp is a not factor of a0 and q not a factor ofan then p_q

is not a zero ofP(x).

2. The test gives a finite set of rational numbers as “candidate zeros”.

3. The test does not give all the zeros of the polynomial, as polynomials may have irrational zeros, and

Example: Factor completely: 2x3_+x2_−13x+6.

Solution:

First, we list all possible rational zeros of the polynomial given by the Rational Zero Test.

p : ±1,±2,±3,±6 q : ±1,±2

p

q : ±1,±2,±3,±6,± 1 2,±

3 2

Use synthetic division to determine which are indeed zeros of the polynomial. Try 1:

2 1 −13 6

1

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution:

First, we list all possible rational zeros of the polynomial given by the Rational Zero Test.

p : ±1,±2,±3,±6 q : ±1,±2

p

q : ±1,±2,±3,±6,± 1 2,±

3 2

Use synthetic division to determine which are indeed zeros of the polynomial. Try 1:

2 1 −13 6

1

2

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution:

First, we list all possible rational zeros of the polynomial given by the Rational Zero Test.

p : ±1,±2,±3,±6 q : ±1,±2

p

q : ±1,±2,±3,±6,± 1 2,±

3 2

Use synthetic division to determine which are indeed zeros of the polynomial. Try 1:

2 1 −13 6

1 2

2

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution:

First, we list all possible rational zeros of the polynomial given by the Rational Zero Test.

p : ±1,±2,±3,±6 q : ±1,±2

p

q : ±1,±2,±3,±6,± 1 2,±

3 2

Use synthetic division to determine which are indeed zeros of the polynomial. Try 1:

2 1 −13 6

1 2

2 3

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution:

First, we list all possible rational zeros of the polynomial given by the Rational Zero Test.

p : ±1,±2,±3,±6 q : ±1,±2

p

q : ±1,±2,±3,±6,± 1 2,±

3 2

Use synthetic division to determine which are indeed zeros of the polynomial. Try 1:

2 1 −13 6

1 2 3

2 3

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution:

First, we list all possible rational zeros of the polynomial given by the Rational Zero Test.

p : ±1,±2,±3,±6 q : ±1,±2

p

q : ±1,±2,±3,±6,± 1 2,±

3 2

Use synthetic division to determine which are indeed zeros of the polynomial. Try 1:

2 1 −13 6

1 2 3

2 3 −10

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution:

First, we list all possible rational zeros of the polynomial given by the Rational Zero Test.

p : ±1,±2,±3,±6 q : ±1,±2

p

q : ±1,±2,±3,±6,± 1 2,±

3 2

Use synthetic division to determine which are indeed zeros of the polynomial. Try 1:

2 1 −13 6

1 2 3 −10

2 3 −10

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution:

First, we list all possible rational zeros of the polynomial given by the Rational Zero Test.

p : ±1,±2,±3,±6 q : ±1,±2

p

q : ±1,±2,±3,±6,± 1 2,±

3 2

Use synthetic division to determine which are indeed zeros of the polynomial. Try 1:

2 1 −13 6

1 2 3 −10

2 3 −10 −4

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution:

First, we list all possible rational zeros of the polynomial given by the Rational Zero Test.

p : ±1,±2,±3,±6 q : ±1,±2

p

q : ±1,±2,±3,±6,± 1 2,±

3 2

Use synthetic division to determine which are indeed zeros of the polynomial. Try -1:

2 1 −13 6

−1

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution:

First, we list all possible rational zeros of the polynomial given by the Rational Zero Test.

p : ±1,±2,±3,±6 q : ±1,±2

p

q : ±1,±2,±3,±6,± 1 2,±

3 2

Use synthetic division to determine which are indeed zeros of the polynomial. Try -1:

2 1 −13 6

−1

2

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution:

First, we list all possible rational zeros of the polynomial given by the Rational Zero Test.

p : ±1,±2,±3,±6 q : ±1,±2

p

q : ±1,±2,±3,±6,± 1 2,±

3 2

Use synthetic division to determine which are indeed zeros of the polynomial. Try -1:

2 1 −13 6

−1 −2

2

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution:

First, we list all possible rational zeros of the polynomial given by the Rational Zero Test.

p : ±1,±2,±3,±6 q : ±1,±2

p

q : ±1,±2,±3,±6,± 1 2,±

3 2

Use synthetic division to determine which are indeed zeros of the polynomial. Try -1:

2 1 −13 6

−1 −2

2 −1

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution:

First, we list all possible rational zeros of the polynomial given by the Rational Zero Test.

p : ±1,±2,±3,±6 q : ±1,±2

p

q : ±1,±2,±3,±6,± 1 2,±

3 2

Use synthetic division to determine which are indeed zeros of the polynomial. Try -1:

2 1 −13 6

−1 −2 1

2 −1

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution:

First, we list all possible rational zeros of the polynomial given by the Rational Zero Test.

p : ±1,±2,±3,±6 q : ±1,±2

p

q : ±1,±2,±3,±6,± 1 2,±

3 2

Use synthetic division to determine which are indeed zeros of the polynomial. Try -1:

2 1 −13 6

−1 −2 1

2 −1 −12

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution:

First, we list all possible rational zeros of the polynomial given by the Rational Zero Test.

p : ±1,±2,±3,±6 q : ±1,±2

p

q : ±1,±2,±3,±6,± 1 2,±

3 2

Use synthetic division to determine which are indeed zeros of the polynomial. Try -1:

2 1 −13 6

−1 −2 1 12

2 −1 −12

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution:

First, we list all possible rational zeros of the polynomial given by the Rational Zero Test.

p : ±1,±2,±3,±6 q : ±1,±2

p

q : ±1,±2,±3,±6,± 1 2,±

3 2

Use synthetic division to determine which are indeed zeros of the polynomial. Try -1:

2 1 −13 6

−1 −2 1 12

2 −1 −12 18

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution (cont.):

Try 2.

2 1 −13 6

2

The remainder is 0; hence, 2 is a zero of the polynomial.

Thus, 2x3+x2−13x+ 6 = (x−2)(2x2+ 5x−3).

The other factor is already quadratic, which we can factor. We get

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution (cont.):

Try 2.

2 1 −13 6

2

2

The remainder is 0; hence, 2 is a zero of the polynomial.

Thus, 2x3+x2−13x+ 6 = (x−2)(2x2+ 5x−3).

The other factor is already quadratic, which we can factor. We get

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution (cont.):

Try 2.

2 1 −13 6

2 4

2

The remainder is 0; hence, 2 is a zero of the polynomial.

Thus, 2x3+x2−13x+ 6 = (x−2)(2x2+ 5x−3).

The other factor is already quadratic, which we can factor. We get

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution (cont.):

Try 2.

2 1 −13 6

2 4

2 5

The remainder is 0; hence, 2 is a zero of the polynomial.

Thus, 2x3+x2−13x+ 6 = (x−2)(2x2+ 5x−3).

The other factor is already quadratic, which we can factor. We get

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution (cont.):

Try 2.

2 1 −13 6

2 4 10

2 5

The remainder is 0; hence, 2 is a zero of the polynomial.

Thus, 2x3+x2−13x+ 6 = (x−2)(2x2+ 5x−3).

The other factor is already quadratic, which we can factor. We get

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution (cont.):

Try 2.

2 1 −13 6

2 4 10

2 5 −3

The remainder is 0; hence, 2 is a zero of the polynomial.

Thus, 2x3+x2−13x+ 6 = (x−2)(2x2+ 5x−3).

The other factor is already quadratic, which we can factor. We get

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution (cont.):

Try 2.

2 1 −13 6

2 4 10 −6

2 5 −3

The remainder is 0; hence, 2 is a zero of the polynomial.

Thus, 2x3+x2−13x+ 6 = (x−2)(2x2+ 5x−3).

The other factor is already quadratic, which we can factor. We get

Example: Factor completely: 2x3_+x2_{−13x+ 6.}

Solution (cont.):

Try 2.

2 1 −13 6

2 4 10 −6

2 5 −3 0

The remainder is 0; hence, 2 is a zero of the polynomial.

Thus, 2x3+x2−13x+ 6 = (x−2)(2x2+ 5x−3).

The other factor is already quadratic, which we can factor. We get

Example

Find the solution set of3x4_{+ 17x}3_{+ 22x}2_{−3x+ 9 = 0.}

Solution:

Letp(x) = 3x4+ 17x3+ 22x2−3x+ 9. Possible Rational Zeros:

p : ±1,±3,±9 q : ±1,±3 p

q : ±1,±3,±9,± 1 3

Try 1.

3 17 22 −3 9

1 3 20 42 39

3 20 42 39 48

Example

Find the solution set of 3x4_{+ 17x}3_{+ 22x}2_{−3x+ 9 = 0.}

Solution (cont.):

Letp(x) = 3x4+ 17x3+ 22x2−3x+ 9. Possible Rational Zeros:

p : ±1,±3,±9 q : ±1,±3 p

q : ±1,±3,±9,± 1 3

Try −3.

3 17 22 −3 9

−3

Example

Find the solution set of 3x4_{+ 17x}3_{+ 22x}2_{−3x+ 9 = 0.}

Solution (cont.):

Letp(x) = 3x4+ 17x3+ 22x2−3x+ 9. Possible Rational Zeros:

p : ±1,±3,±9 q : ±1,±3 p

q : ±1,±3,±9,± 1 3

Try −3.

3 17 22 −3 9

−3

3

Example

Find the solution set of 3x4_{+ 17x}3_{+ 22x}2_{−3x+ 9 = 0.}

Solution (cont.):

Letp(x) = 3x4+ 17x3+ 22x2−3x+ 9. Possible Rational Zeros:

p : ±1,±3,±9 q : ±1,±3 p

q : ±1,±3,±9,± 1 3

Try −3.

3 17 22 −3 9

−3 −9

3

Example

Find the solution set of 3x4_{+ 17x}3_{+ 22x}2_{−3x+ 9 = 0.}

Solution (cont.):

Letp(x) = 3x4+ 17x3+ 22x2−3x+ 9. Possible Rational Zeros:

p : ±1,±3,±9 q : ±1,±3 p

q : ±1,±3,±9,± 1 3

Try −3.

3 17 22 −3 9

−3 −9

3 8

Example

Find the solution set of 3x4_{+ 17x}3_{+ 22x}2_{−3x+ 9 = 0.}

Solution (cont.):

Letp(x) = 3x4+ 17x3+ 22x2−3x+ 9. Possible Rational Zeros:

p : ±1,±3,±9 q : ±1,±3 p

q : ±1,±3,±9,± 1 3

Try −3.

3 17 22 −3 9

−3 −9 −24

3 8

Example

Find the solution set of 3x4_{+ 17x}3_{+ 22x}2_{−3x+ 9 = 0.}

Solution (cont.):

Letp(x) = 3x4+ 17x3+ 22x2−3x+ 9. Possible Rational Zeros:

p : ±1,±3,±9 q : ±1,±3 p

q : ±1,±3,±9,± 1 3

Try −3.

3 17 22 −3 9

−3 −9 −24

3 8 −2

Example

Find the solution set of 3x4_{+ 17x}3_{+ 22x}2_{−3x+ 9 = 0.}

Solution (cont.):

Letp(x) = 3x4+ 17x3+ 22x2−3x+ 9. Possible Rational Zeros:

p : ±1,±3,±9 q : ±1,±3 p

q : ±1,±3,±9,± 1 3

Try −3.

3 17 22 −3 9

−3 −9 −24 6

3 8 −2

Example

Find the solution set of 3x4_{+ 17x}3_{+ 22x}2_{−3x+ 9 = 0.}

Solution (cont.):

Letp(x) = 3x4+ 17x3+ 22x2−3x+ 9. Possible Rational Zeros:

p : ±1,±3,±9 q : ±1,±3 p

q : ±1,±3,±9,± 1 3

Try −3.

3 17 22 −3 9

−3 −9 −24 6

3 8 −2 3

Example

Find the solution set of 3x4_{+ 17x}3_{+ 22x}2_{−3x+ 9 = 0.}

Solution (cont.):

Letp(x) = 3x4+ 17x3+ 22x2−3x+ 9. Possible Rational Zeros:

p : ±1,±3,±9 q : ±1,±3 p

q : ±1,±3,±9,± 1 3

Try −3.

3 17 22 −3 9

−3 −9 −24 6 −9

3 8 −2 3

Example

Find the solution set of 3x4_{+ 17x}3_{+ 22x}2_{−3x+ 9 = 0.}

Solution (cont.):

Letp(x) = 3x4+ 17x3+ 22x2−3x+ 9. Possible Rational Zeros:

p : ±1,±3,±9 q : ±1,±3 p

q : ±1,±3,±9,± 1 3

Try −3.

3 17 22 −3 9

−3 −9 −24 6 −9

3 8 −2 3 0

Example

Find the solution set of3x4_{+ 17x}3_{+ 22x}2_{−3x+ 9 = 0.}

Solution (cont.):

Find the zeros of the other factor 3x3+ 8x2−2x+ 3. Its possible rational zeros are:

p : ±1,±3 q : ±1,±3 p

q : ±1,±3 Try −3.

3 8 −2 3

−3 −9 3 −3

3 −1 1 0

Thus,−3 is a zero of the originalp(x) with multiplicity 2.

Example

Find the solution set of 3x4_{+ 17x}3_{+ 22x}2_{−3x+ 9 = 0.}

Solution (cont.): We have

(x+ 3)(x+ 3)(3x2−x+ 1) = 0.

x=−3 or 3x2−x+ 1 = 0. Using the quadratic formula,

x= −(−1)±

p

(−1)2_−4(3)(1)

2(3) =

1±√11i 6

Therefore, the solution set is

(

−3,1 +

√

11i

6 ,

1−√11i 6

)

Inverse Functions

Let

f

(

x

) = 2

x

+ 1

,

and

g

(

x

) =

1

2

x

−

1

2

.

Observe that

(g◦f)(x) = g(f(x)) =g(2x+ 1) = 1

2(2x+ 1)− 1 2 =x,

(f◦g)(x)=f(g(x)) =f

1 2x−

1 2 = 2 1 2x−

1 2

Inverse Functions

Definition

Two functions

f

and

g

are

inverse functions

of each other if:

(

f

◦

g

)(

x

) =

x

for all

x

∈

dom

g

, AND

(

g

◦

f

)(

x

) =

x

for all

x

∈

dom

f

.

Note:

1.

We write

g

=

f

and

f

=

g

.

2.

(

f

◦

f

)(

x

) =

x

for all

x

∈

dom

f

Example: f(x) = 2x+ 1 andg(x) = 1₂x− 1

2 are inverse

functions of each other.

Thus,f−1_{(x) =} 1 2x−

1 2 and g

−1_{(x) = 2x+ 1.}

One-to-One Functions

Definition

A function

f

is said to be

one-to-one

if

for

all

a, b

∈

dom

f

with

a

6

=

b

,

f

(

a

)

6

=

f

(

b

).

a1

a2

a4

a3

f

b1

b2

b4

b3

one-to-one

b1

b2

b4

b3

a1

a2

a4

a3

inverse mapping

a1

a2

a4

a3

g

b1

b2

b4

b3

not one-to-one

b1

b2

b4

b3

inverse mapping

a1

a2

a4

a3

Theorem

A function

f

has an inverse function if

and only if

f

is a one-to-one function.

a1 a2 a4 a3 f b1 b2 b4 b3 b1 b2 b4 b3

f−1 a1

a2

a4

a3

1.

If

f

(

a

) =

b

, then

f

(

b

) =

a

.

That is (

a, b

)

∈

f

precisely when (

b, a

)

∈

f

.

2.

dom

f

= ran

f

Example: Let f(x) = x2_{+ 1. Doesf have an inverse}

function? Solution:

Is f one-to-one?

Note thatf(1) = 2 andf(−1) = 2. Since two different elements of domf have the same image,f is not one-to-one.

Finding

f

Letf be a one-to-one function. To find f−1_:

Example y=f(x) = 2x+ 1

1. Interchangex and y to

get x=f(y). x= 2y+ 1

2. Solve for y in terms of

x to get y=f−1(x). y= x−1 2

Example: Findg−1 if g(x) = √3 _{x+ 1.} Solution:

y = √3 _{x+ 1} x = √3 _{y+ 1} 3

√

y = x−1 y = (x−1)3

Example: Find the range off(x) = x+ 1

x−2.

Solution: Note that ranf = domf−1. We obtain f−1 first:

y = x+ 1 x−2

x = y+ 1 y−2 x(y−2) = y+ 1

xy−2x = y+ 1 xy−y = 2x+ 1 y(x−1) = 2x+ 1

y = 2x+ 1 x−1

f−1_{(x) =} 2x+ 1

x−1

Example: Let f(x) = x2+ 1. Doesf have an inverse function?

Another Solution: We can try and find f−1(x):

y=x2+ 1 x=y2+ 1 y2 =x−1

y=±√x−1

The last equation does not represent a function.

Horizontal Line Test

Theorem

A function

f

is one-to-one if and only if

any horizontal line intersects the graph of

f

in at most one point.

Examples:

f(x) = x+ 1

one-to-one

f(x) =|x|

not one-to-one

f(x) = √3 _x

Example: Let f(x) = x2_{+ 1. Doesf have an inverse}

function?

Another Solution:

Is f one-to-one?

The graph off(x) =x2+ 1 is an upward parabola with vertex at (0,1), it fails the horizontal line test. Hence,f is not one-to-one.

Recall: If

f

exists, then (

a, b

)

∈

f

precisely

when (

b, a

)

∈

f

(a, b)

(b, a)

(a+b 2 ,

b+a 2 )

slope b_a−₋a_b slope−1

`⊥: slope 1

y−b+a

2 = 1(x− a+b

2 )

y=x

y=f(x)

The graph of

f

is the reflection of the graph of

f

about the line with equation

y

=

x

.

f(x) = 2x+ 1 f−1(x) = 1₂x− 1

2

The graph of

f

is the reflection of the graph of

f

about the line with equation

y

=

x

.

f(x) = 2x+ 1 f−1(x) = 1₂x− 1

2

Restriction of Domain

Iff is not one-to-one, we may restrict its domain such that f is one-to-one on the restriction. The resulting function has an inverse.

Example: f(x) =x2+ 1.

Since

f is not one-to one on

R,

we restrict domf to [0,∞).

x=y2_{+ 1, y≥0}

y2 =x−1,y ≥0 y=±√x−1,y ≥0 y =√x−1 since y∈[0,∞)

Restriction of Domain

Iff is not one-to-one, we may restrict its domain such that f is one-to-one on the restriction. The resulting function has an inverse.

Example: f(x) =x2+ 1.

Since f is not one-to one on

R, we restrict domf to

[0,∞).

x=y2+ 1, y≥0 y2 _{=x−1,y ≥0}

y=±√x−1,y≥0 y=√x−1 since y∈[0,∞)

Restriction of Domain

Iff is not one-to-one, we may restrict its domain such that f is one-to-one on the restriction. The resulting function has an inverse.

Example: f(x) =x2+ 1.

Since f is not one-to one on

R, we restrict domf to

[0,∞).

x=y2+ 1, y≥0 y2 _{=x−1,y ≥0}

y=±√x−1,y≥0 y=√x−1 since y∈[0,∞)

Recap:

Relate the zeros of a polynomial with solving equations and factoring

Use the rational zero test to list down possible zeros of a polynomial

Determine the inverse of a function (if it exists)

Recognize one-to-one functions

Exercises:

1. Solve for all (including irrational and complex roots if any)

xin the equation 27x4+ 27x3−6x2−2x+ 4 = 0.

2. Determine if each function is one-to-one. If not, explain why.

1. {(2,3),(3,5),(5,7)} 2. {(x, y) |y=|x+ 2|}

3. Determine whether the given pair are inverses of each other.

1. f(x) = 20−5x,

g(x) =−0.2x+ 4

2. f(x) = _x−1₃,

g(x) = 3 + 1_x

4. Give the domain and range of the following functions and their inverses.