copyright © 2007 pearson education, inc. slide 4-1

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Page 1: Copyright © 2007 Pearson Education, Inc. Slide 4-1

Copyright © 2007 Pearson Education, Inc. Slide 4-1

Page 2: Copyright © 2007 Pearson Education, Inc. Slide 4-1

Copyright © 2007 Pearson Education, Inc. Slide 4-2

Chapter 4: Rational, Power, and Root Functions

4.1 Rational Functions and Graphs

4.2 More on Graphs of Rational Functions

4.3 Rational Equations, Inequalities, Applications, and Models

4.4 Functions Defined by Powers and Roots

4.5 Equations, Inequalities, and Applications Involving Root Functions

Page 3: Copyright © 2007 Pearson Education, Inc. Slide 4-1

Copyright © 2007 Pearson Education, Inc. Slide 4-3

Note: This property does not say that the two equations are equivalent. The new equation may have more solutions than the original.

e.g.

4.5 Equations, Inequalities, and Applications Involving Root Functions

Power Property

If P and Q are algebraic expressions, then every solution of the equation P = Q is also a solution of the equation Pn = Qn, for any positive integer n.

2 4 with compared2 2

xxx

Page 4: Copyright © 2007 Pearson Education, Inc. Slide 4-1

Copyright © 2007 Pearson Education, Inc. Slide 4-4

4.5 Solving Equations Involving Root Functions

1. Isolate a term involving a root on one side of the equation.

2. Raise both sides of the equation to a power that will eliminate the radical or rational exponent.

3. Solve the resulting equation. (If a root is still present after Step 2, repeat Steps 1 and 2.)

4. Check each proposed solution in the original equation.

Page 5: Copyright © 2007 Pearson Education, Inc. Slide 4-1

Copyright © 2007 Pearson Education, Inc. Slide 4-5

4.5 Solving an Equation Involving Square Roots

Example Solve

Analytic Solution

.111 xx

111 xx

Isolate the radical.xx 111

Square both sides. 22111 xx

22111 xxx Write in standard form and solve.

2or5)2)(5(0

1030 2

xxxx

xx

Page 6: Copyright © 2007 Pearson Education, Inc. Slide 4-1

Copyright © 2007 Pearson Education, Inc. Slide 4-6

4.5 Solving an Equation Involving Square Roots

These solutions must be checked in the original equation.

value.an is 5 So False.19?1516?1)5()5(11.5Let

extraneous

x

solution.only theis 2 So True.11?129?12211

.2Let

x

Page 7: Copyright © 2007 Pearson Education, Inc. Slide 4-1

Copyright © 2007 Pearson Education, Inc. Slide 4-7

4.5 Solving an Equation Involving Square Roots

Graphical Solution The equation in the second step

of the analytic solution has the same solution set asthe original equation. Graphand solve y1 = y2.

The only solution is at x = 2.

xyxy 1 and11 21

Page 8: Copyright © 2007 Pearson Education, Inc. Slide 4-1

Copyright © 2007 Pearson Education, Inc. Slide 4-8

4.5 Solving an Equation Involving Cube Roots

Example Solve

Solution

.53 33 2 xx

2293

)1(2)5)(1(433

05353

53

2

2

2

3333 2

33 2

53

x

x

xxxx

xx

xx

. isset solution theshown that becan It 2293-

Page 9: Copyright © 2007 Pearson Education, Inc. Slide 4-1

Copyright © 2007 Pearson Education, Inc. Slide 4-9

4.5 Solving an Equation Involving Roots (Squaring Twice)

Example Solve

Solution

3or10)1)(3(032

)1(412

121112132

11321132

2

2

22

22

121

1132

xxxxxx

xxx

xxxxx

xxxx

xx

xx

A check shows that –1 and 3 are solutions of the original equation.

Isolate radical.

Square both sides.

Isolate radical.

Square both sides.

Write in standard quadratic form and solve.

.1132 xx

Page 10: Copyright © 2007 Pearson Education, Inc. Slide 4-1

Copyright © 2007 Pearson Education, Inc. Slide 4-10

4.5 Solving Inequalities Involving Rational Exponents

Example Solve the inequality

Solution The associated equation solution in the previous example was Let

.53 33 2 xx

. and 2293

2293

.graph and 53 33 2 yxxy

Use the x-intercept method to solve this inequality and determine the interval where the graph lies below the x-axis. The solution is the interval . 2

2932

293 ,

Page 11: Copyright © 2007 Pearson Education, Inc. Slide 4-1

Copyright © 2007 Pearson Education, Inc. Slide 4-11

4.5 Application: Solving a Cable Installation Problem

A company wishes to run a utility cable from point A on the shore to an installation at point B on the island (see figure). The island is 6 miles from shore. It costs $400 per mile to runcable on land and $500 per mile underwater. Assume that the cable starts at point A and runs along the shoreline, then angles and runs underwater to the island. Let x represent the distance from C at which the underwater portion of the cable run begins, and the distance between A and C be 9 miles.

Page 12: Copyright © 2007 Pearson Education, Inc. Slide 4-1

Copyright © 2007 Pearson Education, Inc. Slide 4-12

4.5 Application: Solving a Cable Installation Problem

(a) What are the possible values of x in this problem?

(b) Express the cost of laying the cable as a function of x.

(c) Find the total cost if three miles of cable are on land.

(d) Find the point at which the line should begin to angle in order to minimize the total cost. What is this total cost?

Solution

(a) The value of x must be real where

(b) Let k be the length underwater. Using the Pythagorean theorem,

.90 x

.0,36

62

222

kxk

xk

Page 13: Copyright © 2007 Pearson Education, Inc. Slide 4-1

Copyright © 2007 Pearson Education, Inc. Slide 4-13

4.5 Application: Solving a Cable Installation Problem

The cost of running cable is price miles. If C is the total cost (in dollars) of laying cable across land and underwater, then

(c) If 3 miles of cable are on land, then 3 = 9 – x, giving

x = 6.

.36500)9(400)( 2xxxC

dollars 64.5442

636500)69(400)6( 2

C

Page 14: Copyright © 2007 Pearson Education, Inc. Slide 4-1

Copyright © 2007 Pearson Education, Inc. Slide 4-14

4.5 Application: Solving a Cable Installation Problem

(d) Using the graphing calculator, find the minimum value of y1 = C(x) on the interval (0,9].

The minimum value of the function occurs when x = 8. So 9 – 8 = 1 mile should be along land, and

miles underwater. The cost is

08.6136 2

dollars. 5400836500)89(400)8( 2 C