Questions and Answers #4 Piecewise-Defined Functions

Questions and Answers Sheet 4 Piecewise-Defined Functions Question #1 Solve the absolute value equation or indicate that the equation has no solution. |x+6|=5 (a) {1} (b) {-11, -1} (c) {11, -1} (d) ∅ Answer: Step 1 To solve: The absolute value function |x+6|=5. Concept used: Absolute value function: |u|=a, a>0 The solutions are u=-a and u=a. Step 2 Calculation: If |x+6|=5 then x+6=-5 and x+6=5. Solve the equation x+6=-5 as follows: x+6=-5 x=-5-6 x=-11 Solve the equation x+6=5 as follows: x+6=5 x=5-6 x=-1 Therefore, the solution of |x+6|=5 is {-11,-1}. Thus, option (b) given as {-11,-1} is correct answer. Question #2 In the following, simplify using absolute values as necessary. 4 √𝑚8 Answer: 4 Given information: An expression is given as √𝑚8 . 4 Calculations: We have been given an expression is √𝑚8 . For finding the odd and even roots by using the absolute value as necessary, we must know about the power property of exponent i.e. (𝑎𝑚 )𝑛 = 𝑎𝑚∗𝑛 . 4 ⇒ √𝑚8 [Simplify the fourth root of 𝑚8 ] 4 ⇒ √(𝑚2 )4 [Break the power exponent in the form of (𝑎𝑚 )𝑛 = 𝑎𝑚∗𝑛 ] 𝑛 ⇒ |𝑚2 | [ √𝑎𝑛 = |𝑎|, if the value of n is even.] 4 Thus, index value of √𝑚8 is even. 4 Hence, the simplification of √𝑚8 is |𝑚2 | Question #3 In the following, simplify using absolute values as necessary. 5 √𝑛20 Answer: 5 Given information: An expression is given as √𝑛20 . 5 Calculations: We have been given an expression is √𝑛20 . For finding the odd and even roots by using the absolute value as necessary, we must know about the power property of exponent i.e. (𝑎𝑚 )𝑛 = 𝑎𝑚∗𝑛 . 5 ⇒ √𝑛20 [Simplify the fifth root of 𝑛20 ] 5 ⇒ √(𝑛4 )5 [Break the power exponent in the form of (𝑎𝑚 )𝑛 = 𝑎𝑚∗𝑛 ] 𝑛 ⇒ 𝑛4 [ √𝑎𝑛 = |𝑎|, if the value of n is even.] 5 Thus, index value of √𝑛20 is even. 5 Hence, the simplification of √𝑛20 is 𝑛4 Question #4 In the following, simplify using absolute values as necessary. 3 √216𝑎6 Answer: 3 Given information: An expression is given as √216𝑎6 . 3 Calculations: We have been given an expression is √216𝑎6 . For finding the odd and even roots by using the absolute value as necessary, we must know about the power property of exponent i.e. (𝑎𝑚 )𝑛 = 𝑎𝑚∗𝑛 3 ⇒ √216𝑎6 [Simplify the cube root of 216𝑎6 ] 3 ⇒ √(6𝑎2 )3 [Break the power exponent in the form of (𝑎𝑚 )𝑛 = 𝑎𝑚∗𝑛 ] 𝑛 ⇒ 6𝑎2 [ √𝑎𝑛 = 𝑎, if the value of n is odd.] 3 Thus, index value of √216𝑎6 is odd. 3 Hence, the simplification of √216𝑎6 is 6𝑎2 Question #5 In the following, simplify using absolute values as necessary. 5 √32𝑏 20 Answer: 5 Given information: An expression is given as √32𝑏 20 . 5 Calculations: We have been given an expression is √32𝑏 20 . For finding the odd and even roots by using the absolute value as necessary, we must know about the power property of exponent i.e. (𝑎𝑚 )𝑛 = 𝑎𝑚∗𝑛 5 ⇒ √32𝑏 20 [Simplify the cube root of 32𝑏 20 ] 5 ⇒ √(2𝑏 4 )5 [Break the power exponent in the form of (𝑎𝑚 )𝑛 = 𝑎𝑚∗𝑛 ] 𝑛 ⇒ 2𝑏 4 [ √𝑎𝑛 = 𝑎, if the value of n is odd.] 5 Thus, index value of √32𝑏 20 is odd. 5 Hence, the simplification of √32𝑏 20 is 2𝑏 4 Question #6 Describe all numbers x that are at a distance of 1 2 from the number -4. Express this using absolute value notation. Answer: Given: Number x at a distance of 1 2 from the number -4. Calculation: The given statement is Number x at distance of Number x at a distance of 𝑦 = |𝑥| + 1 2 Number x at a distance of 𝑦 = |𝑥 − (−4)| + 𝑦 = |𝑥 + 4| + 1 1 1 1 2 from the number -4. 2 1 2 from the number -4. 2 2 Conclusion: 1 Thus, the absolute value equation is 𝑦 = |𝑥 + 4| + 2 Question #7 In the following, simplify using absolute values as necessary. √𝑎14 Answer: Given information: An expression is given as √𝑎14 . Calculations: We have been given an expression is √𝑎14 . For finding the odd and even roots by using the absolute values as necessary, we must know about the power property of exponent i.e. (𝑎𝑚 )𝑛 = 𝑎𝑚∗𝑛 . ⇒ √𝑎14 [Simplify the square root of 𝑎14 ] ⇒ √(𝑎7 )2 [Break the power exponent in the form of (𝑎𝑚 )𝑛 = 𝑎𝑚∗𝑛 ] 𝑛 ⇒ |𝑎3 | [ √𝑎𝑛 = |𝑎|, if the value of n is even.] Thus, index value of √𝑎14 is even. Hence, the simplification of √𝑎14 is |𝑎3 |. Question #8 In the following, simplify using absolute values as necessary. √𝑤 24 Answer: Given information: An expression is given as √𝑤 24 . Calculations: We have been given an expression is √𝑤 24 . For finding the odd and even roots by using the absolute values as necessary, we must know about the power property of exponent i.e. (𝑎𝑚 )𝑛 = 𝑎𝑚∗𝑛 . ⇒ √𝑤 24 [Simplify the square root of 𝑤 24 ] ⇒ √(𝑤 12 )2 [Break the power exponent in the form of (𝑎𝑚 )𝑛 = 𝑎𝑚∗𝑛 ] 𝑛 ⇒ |𝑤 12 | [ √𝑎𝑛 = |𝑎|, if the value of n is even.] Thus, index value of √𝑤 24 is even. Hence, the simplification of √𝑤 24 is |𝑤 12 |. Question #9 In the following, simplify using absolute value signs as needed. √96𝑟 3 𝑠 3 Answer: Given information: An expression is given as √96𝑟 3 𝑠 3 . Calculations: We have been given an expression is √96𝑟 3 𝑠 3 . To simplify the radical expressions by using absolute value as needed, we must know about the radical properties and the index value of root is even or odd as shown below: 𝑛 We know that, (𝑎𝑏)𝑚 = 𝑎𝑚 ∗ 𝑏 𝑚 and the corresponding product property of radical expression is √𝑎𝑏 = 𝑛 𝑛 √𝑎 ∗ √𝑏. ⇒ √96𝑟 3 𝑠 3 [Simplify the square root of 96𝑟 3 𝑠 3 with radical properties] ⇒ √16𝑟 2 ∗ 𝑠 2 ∗ 6𝑟𝑠 [Rewrite 96𝑟 3 𝑠 3 = 16𝑟 2 ∗ 𝑠 2 ∗ 6𝑟𝑠 ] 𝑛 𝑛 𝑛 ⇒ √(4𝑟𝑠)2 ∗ √6𝑟𝑠 [By using radical property √𝑎𝑏 = √𝑎 ∗ √𝑏 ] 𝑛 ⇒ 2|𝑟𝑠|√6𝑟𝑠 [ √𝑎𝑛 = |𝑎|, if the value of index is even] Hence, the simplification of √96𝑟 3 𝑠 3 is 2|𝑟𝑠|√6𝑟𝑠 Question #10 In the following, simplify using absolute value signs as needed. 3 √80𝑥 7 𝑦 6 Answer: 3 Given information: An expression is given as √80𝑥 7 𝑦 6 . 3 Calculations: We have been given an expression is √80𝑥 7 𝑦 6 . To simplify the radical expressions by using absolute value as needed, we must know about the radical properties and the index value of root is even or odd as shown below: 𝑛 We know that, (𝑎𝑏)𝑚 = 𝑎𝑚 ∗ 𝑏 𝑚 and the corresponding product property of radical expression is √𝑎𝑏 = 𝑛 𝑛 √𝑎 ∗ √𝑏. 3 ⇒ √80𝑥 7 𝑦 6 [Simplify the cube root of 80𝑥 7 𝑦 6 with radical properties] 3 ⇒ √8𝑥 6 ∗ 𝑦 6 ∗ 10𝑥 [Rewrite 80𝑥 7 𝑦 6 = 8𝑥 6 ∗ 𝑦 6 ∗ 10𝑥 ] 𝑛 𝑛 3 3 𝑛 ⇒ √(2𝑥 2 𝑦 2 )3 ∗ √10𝑥 [By using radical property √𝑎𝑏 = √𝑎 ∗ √𝑏 ] 3 𝑛 ⇒ 2𝑥 2 𝑦 2 ∗ √10𝑥 [ √𝑎𝑛 = |𝑎|, if the value of index is odd] 3 ⇒ 2𝑥 2 𝑦 2 √10𝑥 3 3 Hence, the simplification of √80𝑥 7 𝑦 6 is 2𝑥 2 𝑦 2 √10𝑥