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PLEASE COMPLETE THE PREREQUISITE SKILLS PG 412 #1-12 CHAPTER 6: RATIONAL EXPONENTS AND RADICAL FUNCTIONS BIG IDEAS: ο§ USE RATIONAL EXPONENTS ο§ PERFORMING FUNCTION OPERATIONS AND FINDING INVERSE FUNCTIONS β’ SOLVING RADICAL EQUATIONS LESSON 1: EVALUATE NTH ROOTS AND USE RATIONAL EXPONENTS WHAT IS THE RELATIONSHIP BETWEEN NTH ROOTS AND RATIONAL EXPONENTS? ESSENTIAL QUESTION VOCABULARY β’ Nth root of a: For an integer n greater than 1, if bn = a, then b is an nth root of a. written as π π β’ Index of a radical: The integer n, greater than 1, in the expression π π EXAMPLE 1 Find nth roots Find the indicated real nth root(s) of a. a. n = 3, a = β216 b. n = 4, a = 81 SOLUTION a. Because n = 3 is odd and a = β216 < 0, β216 has one real cube root. Because (β6)3 = β216, you can write = 3β β216 = β6 or (β216)1/3 = β6. b. Because n = 4 is even and a = 81 > 0, 81 has two real fourth roots. Because 34 = 81 and (β3)4 = 81, you can write ±4β 81 = ±3 Evaluate expressions with rational exponents EXAMPLE 2 Evaluate (a) 163/2 and (b) 32β3/5. SOLUTION Rational Exponent Form = (161/2)3 = 43 = 64 a. 163/2 b. 32β3/5 = = 1 = 323/5 1 23 = 1 (321/5)3 1 8 Radical Form 3 163/2 = (ο 16 ) 32β3/5 = = 1 323/5 1 = 23 = = 43 = 64 1 (ο5 32 )3 1 8 EXAMPLE 3 Approximate roots with a calculator Keystrokes Expression a. 91/5 b. c. ( Display 9 1 5 1.551845574 123/8 12 3 8 2.539176951 4 7 3 4 4.303517071 ο 7)3 = 73/4 for Examples 1, 2 and 3 GUIDED PRACTICE Find the indicated real nth root(s) of a. 1. n = 4, a = 625 SOLUTION 2. 3. ±5 n = 6, a = 64 SOLUTION SOLUTION 4. ±2 n = 3, a = β64. β4 n = 5, a = 243 SOLUTION 3 for Examples 1, 2 and 3 GUIDED PRACTICE Evaluate expressions without using a calculator. 5. 45/2 SOLUTION 6. 7. 813/4 32 9β1/2 SOLUTION SOLUTION 27 8. 17/8 1 3 SOLUTION 1 GUIDED PRACTICE for Examples 1, 2 and 3 Evaluate the expression using a calculator. Round the result to two decimal places when appropriate. Expression 9. 42/5 SOLUTION 1.74 10. β 64 2/3 SOLUTION 0.06 11. (4β 16)5 SOLUTION 32 12. (3β β30)2 SOLUTION 9.65 EXAMPLE 4 Solve equations using nth roots Solve the equation. a. 4x5 = 128 x5 = 32 Divide each side by 4. 5 32 x = ο Take fifth root of each side. x = 2 Simplify. Solve equations using nth roots EXAMPLE 4 b. (x β 3)4 = 21 4 xβ3 = + β ο 21 Take fourth roots of each side. 4 x = + β ο21 + 3 x = 4 x 5.14 ο21 + 3 or Add 3 to each side. or x x = β4ο21 + 3 0.86 Write solutions separately. Use a calculator. WHAT IS THE RELATIONSHIP BETWEEN NTH ROOTS AND RATIONAL EXPONENTS? The nth root of a can be written as a to the 1 π ESSENTIAL QUESTION SIMPLIFY THE EXPRESSION: 43*48 LESSON 2: APPLY PROPERTIES OF RATIONAL EXPONENTS HOW ARE THE PROPERTIES OF RATIONAL EXPONENTS RELATED TO PROPERTIES OF INTEGER EXPONENTS? ESSENTIAL QUESTION VOCABULARY β’ Simplest form of a radical: A radical with index n is in simplest form if the radicand has no perfect nth powers as factors and any denominator has been rationalized β’ Like radicals: Radical expressions with the same index and radicand Use properties of exponents EXAMPLE 1 Use the properties of rational exponents to simplify the expression. a. 71/4 71/2 b. (61/2 41/3)2 c. (45 35)β1/5 d. e. 5 51/3 = = (61/2)2 (41/3)2 = [(4 3)5]β1/5 51 51/3 421/3 2 61/3 = 73/4 = 7(1/4 + 1/2) = = 5(1 β 1/3) 42 6 1/3 = 6(1/2 2) 4(1/3 2) = (125)β1/5 = 12[5 (β1/5)] = 52/3 2 = (71/3)2 = 7(1/3 2) = 72/3 = 61 42/3 = 6 42/3 = 12 β1 = 1 12 Use properties of radicals EXAMPLE 3 Use the properties of radicals to simplify the expression. a. b. 12 3 4 4 80 5 3 18 = = 4 3 12 18 80 5 = 4 = 16 3 = 216 = 6 2 Product property Quotient property Write radicals in simplest form EXAMPLE 4 Write the expression in simplest form. a. 3 ο 135 3 = ο 27 5 = ο 27 3 3 = 3 3ο 5 ο 5 Factor out perfect cube. Product property Simplify. Write radicals in simplest form EXAMPLE 4 5 b. ο 7 5 ο 8 = 5 ο 7 5 5 5 ο 8 ο 4 ο 4 Make denominator a perfect fifth power. 5 = ο 28 Product property 5 ο 32 5 = ο 28 2 Simplify. Add and subtract like radicals and roots EXAMPLE 5 Simplify the expression. 4 4 4 4 a. ο 10 + 7 ο 10 = (1 + 7) ο 10 = 8 ο 10 b. 2 (81/5) + 10 (81/5) = (2 +10) c. 3 3 3 ο 54 β ο 2 = ο 27 (81/5) = 12 (81/5) 3 3 3 3 3 ο 2 β ο 2 = 3 ο 2 β ο 2 = (3 β 1) ο 2 = 2 ο 2 3 for Examples 3, 4, and 5 GUIDED PRACTICE Simplify the expression. 6. 4 27 4 3 8. 5 3 4 5 SOLUTION 7. 3 ο 24 SOLUTION 2 3 ο 250 9. 3 ο 5 3 + ο 40 3 ο 2 SOLUTION 5 SOLUTION 3 3ο 5 EXAMPLE 6 Simplify expressions involving variables Simplify the expression. Assume all variables are positive. 3 a. 3 b. (27p3q12)1/3 = 271/3(p3)1/3(q12)1/3 c. d. 3 ο 64y6 = ο 43(y2)3 4 m4 n8 4 = 14xy 1/3 2x 3/4 z β6 ο 4 ο m4 n8 4 = 3 = ο 43 ο m4 4 ο (n2)4 = = 7x(1 β 3/4)y1/3z β(β6) ο (y2)3 = 3p(3 = 4y2 1/3)q(12 m n2 = 7x1/4y1/3z6 1/3) = 3pq4 EXAMPLE 7 Write variable expressions in simplest form Write the expression in simplest form. Assume all variables are positive. a. 5 ο 4a8b14c5 5 = ο 4a5a3b10b4c5 5 = ο a5b10c5 b. 3 x y8 = = 3 5 = ab2c ο 4a3b4 x y y8 y 3 5 ο 4a3b4 Product property Simplify. Make denominator a perfect cube. xy y9 Factor out perfect fifth powers. Simplify. EXAMPLE 7 Write variable expressions in simplest form 3 = ο xy 3 ο y9 Quotient property 3 = οxy y3 Simplify. EXAMPLE 8 Add and subtract expressions involving variables Perform the indicated operation. Assume all variables are positive. a. b. c. 1 3 οw + οw = 5 5 3xy1/4 β 8xy1/4 3 1 + 5 5 = (3 β 8) xy1/4 οw = 4 οw 5 = β5xy1/4 3 3 3 3 2 z 5 β ο ο 54z = 12z ο 2z2 β 3z ο 2z2 12 2z 3 = (12z β 3z) ο 2z2 3 = 9z ο 2z2 for Examples 6, 7, and 8 GUIDED PRACTICE Simplify the expression. Assume all variables are positive. 10. 3 ο 27q9 SOLUTION 11. 5 12. 3q3 x10 3x 1/2 y 1/2 SOLUTION 13. y5 SOLUTION 6xy 3/4 x2 y 2x1/2y1/4 ο 9w5 β w ο w3 SOLUTION 2w2 ο w HOW ARE THE PROPERTIES OF RATIONAL EXPONENTS RELATED TO PROPERTIES OF INTEGER EXPONENTS? All properties of integer exponents also apply to rational exponents ESSENTIAL QUESTION LET F(X) = 3X + 5. FIND F(-6) LESSON 3 PERFORM FUNCTION OPERATIONS AND COMPOSITION WHAT OPERATIONS CAN BE PERFORMED ON A PAIR OF FUNCTIONS TO OBTAIN A THIRD FUNCTION? ESSENTIAL QUESTION VOCABULARY β’ Power Function: A function of the form y=axb, where a is a real number and b is a rational number β’ Composition: The composition of a function g with a function f is h(x) = f(f(x)). EXAMPLE 1 Add and subtract functions Let f (x) = 4x1/2 and g(x) = β9x1/2. Find the following. a. f(x) + g(x) SOLUTION f (x) + g(x) b. = 4x1/2 + (β9x1/2) = [4 + (β9)]x1/2 = β5x1/2 f(x) β g(x) SOLUTION f (x) β g(x) = 4x1/2 β (β9x1/2) = [4 β (β9)]x1/2 = 13x1/2 EXAMPLE 1 c. Add and subtract functions the domains of f + g and f β g SOLUTION The functions f and g each have the same domain: all nonnegative real numbers. So, the domains of f + g and f β g also consist of all nonnegative real numbers. Multiply and divide functions EXAMPLE 2 Let f (x) = 6x and g(x) = x3/4. Find the following. a. f (x) g(x) SOLUTION f (x) b. g(x) = (6x)(x3/4) = 6x(1 + 3/4) = 6x7/4 f (x) g(x) SOLUTION f (x) g(x) = 6x x3/4 = 6x(1 β 3/4) = 6x1/4 EXAMPLE 2 c. Multiply and divide functions the domains of f g and f g SOLUTION The domain of f consists of all real numbers, and the domain of g consists of all nonnegative real numbers. So, the domain of f g consists of all nonnegative real numbers. Because g(0) = 0, the domain of is restricted to all positive real numbers. f g EXAMPLE 3 Solve a multi-step problem Rhinos For a white rhino, heart rate r (in beats per minute) and life span s (in minutes) are related to body mass m (in kilograms) by these functions: r(m) = 241mβ0.25 β’ β’ s(m) = (6 Find r(m) s(m). Explain what this product represents. 106)m0.2 EXAMPLE 3 Solve a multi-step problem SOLUTION STEP 1 Find and simplify r(m) s(m). r(m) s(m) = 241m β0.25 [ (6 106)m0.2 ] = 241(6 106)m(β0.25 + 0.2) = (1446 106)m = (1.446 109)m β0.05 β0.05 Write product of r(m) and s(m). Product of powers property Simplify. Use scientific notation. EXAMPLE 3 Solve a multi-step problem STEP 2 Interpret r(m) s(m). Multiplying heart rate by life span gives the total number of heartbeats for a white rhino over its entire lifetime. GUIDED PRACTICE for Examples 1, 2, and 3 Let f (x) = β2x2/3 and g(x) = 7x2/3. Find the following. 1. f (x) + g(x) SOLUTION f (x) + g(x) 2. = β2x2/3 + 7x2/3 = (β2 + 7)x2/3 = β2x2/3 β 7x2/3 = [β2 + ( β7)]x2/3 = 5x2/3 f (x) β g(x) SOLUTION f (x) β g(x) = β9x2/3 GUIDED PRACTICE 3. for Examples 1, 2, and 3 the domains of f + g and f β g SOLUTION all real numbers; all real numbers for Examples 1, 2, and 3 GUIDED PRACTICE Let f (x) = 3x and g(x) = x1/5. Find the following. 4. f (x) SOLUTION 5. g(x) 3x6/5 f (x) g(x) SOLUTION 3x4/5 GUIDED PRACTICE 6. the domains of f for Examples 1, 2, and 3 g and f g SOLUTION all real numbers; all real numbers except x=0. GUIDED PRACTICE for Examples 1, 2, and 3 Rhinos 7. Use the result of Example 3 to find a white rhinoβs number of heartbeats over its lifetime if its body mass is 1.7 105 kilograms. SOLUTION about 7.92 108 heartbeats WHAT OPERATIONS CAN BE PERFORMED ON A PAIR OF FUNCTIONS TO OBTAIN A THIRD FUNCTION? Two functions can be combined by the operations: +, -, x, ÷ and composition ESSENTIAL QUESTION SOLVE X=4Y3 FOR Y LESSON 4: USE INVERSE FUNCTIONS HOW DO YOU FIND AN INVERSE RELATION OF A GIVEN FUNCTION? ESSENTIAL QUESTION VOCABULARY β’ Inverse relation: A relation that interchanges the input and output values of the original relation. The graph of an inverse relation is a reflection of the graph of the original relation, with y=x as the line of reflection β’ Inverse function: An inverse relation that is a function. Functions f and g are inverses provided that f(g(x)) = x and g(f(x)) = x EXAMPLE 1 Find an inverse relation Find an equation for the inverse of the relation y = 3x β 5. y = 3x β 5 x = 3y β 5 x + 5 = 3y 1 5 x+ =y 3 3 Write original relation. Switch x and y. Add 5 to each side. Solve for y. This is the inverse relation. Verify that functions are inverses EXAMPLE 2 Verify that f(x) = 3x β 5 and f are inverse functions. β1(x) 5 1 x+ 3 3 = SOLUTION STEP 1 STEP 2 Show: that f(f β1(x)) = x. Show: that f β1(f(x)) = x. f (f β1(x)) =f =3 1 x+ 3 5 3 1 x+ 3 5 3 f β1(f(x)) = β5 f β1((3x β 5) = 1 (3x β 5) + 3 =x+5β5 =xβ =x =x 5 5 + 3 3 5 3 for Examples 1, 2, and 3 GUIDED PRACTICE Find the inverse of the given function. Then verify that your result and the original function are inverses. 1. f(x) = x + 4 SOLUTION 2. 3. xο1 xβ4 =y SOLUTION f(x) = 2x β 1 SOLUTION x+1 2 f(x) = β3x β 1 =y ο3 =y for Examples 1, 2, and 3 GUIDED PRACTICE 4. Fitness: Use the inverse function in Example 3 to find the length at which the band provides 13 pounds of resistance. SOLUTION 48 inches HOW DO YOU FIND AN INVERSE RELATION OF A GIVEN FUNCTION? Write the original equation. Switch x and y. Solve for y. ESSENTIAL QUESTION EXPAND AND SOLVE: (X-5)2 LESSON 6: SOLVE RADICAL EQUATIONS WHY IS IT NECESSARY TO CHECK EVERY APPARENT SOLUTION OF A RADICAL EQUATION IN THE ORIGINAL EQUATION? ESSENTIAL QUESTION VOCABULARY β’ Radical equation: An equation with one or more radicals that have variables in their radicands β’ Extraneous solution: An apparent solution that must be rejects because it does not satisfy the original equation. Solve a radical equation EXAMPLE 1 Solve 3 ο 2x+7= 3. 3 ο 2x+7 ( ο3 2x+7 )3 =3 Write original equation. = 33 Cube each side to eliminate the radical. 2x+7 = 27 2x = 20 x = 10 Simplify. Subtract 7 from each side. Divide each side by 2. Solve a radical equation EXAMPLE 1 CHECK Check x = 10 in the original equation. 3 ο 2(10)+7 3 ο 27 3 =? 3 Substitute 10 for x. =? 3 Simplify. = 3 Solution checks. GUIDED PRACTICE for Example 1 Solve equation. Check your solution. 1. 3β x β 9 = β1 ANSWER 2. 3. x = 512 (ο x+25 ) = 4 ANSWER x = β9 (23ο x β 3 ) = 4 ANSWER x = 11 WHY IS IT NECESSARY TO CHECK EVERY APPARENT SOLUTION OF A RADICAL EQUATION IN THE ORIGINAL EQUATION? Raising both sides of an equation to the same power sometimes results in an extraneous solution ESSENTIAL QUESTION