Questions and Answers #12 Modeling

Questions and Answers Sheet 12 Modeling Question #1 a) Domain and range of f b) The table for 𝑓 − 1(𝑥) and the domain and range. x 1 10 100 2 . Given: f ( x ) 0 1 Answer: Formula used: The values of x and f(x) interchange for the table of inverse function. Calculations: x 1 10 100 2 a) f ( x ) 0 1 Here the domain of 𝑓 = {1,10,100} The range of 𝑓 = {0,1,2} b) We get the table for the inverse by interchanging the rows of x and f(x). We get x 0 1 2 f − 1 1 10 100 The domain for 𝑓 − 1(𝑥) = {0,1,2} The range for 𝑓 − 1(𝑥) = {1,10,100} Conclusion: a) Here the domain of 𝑓 = {1,10,100} . The range of 𝑓 = {0,1,2} x 0 1 2 b) f − 1( x) 1 10 100 The domain for 𝑓 − 1(𝑥) = {0,1,2} The range for 𝑓 − 1(𝑥) = {1,10,100} Question #2 The value of 𝑓 −1 (0) using the table given. Given: a table of values for the functions f(x) and g(x). x 0 2 4 6 8 f ( x) 8 2 0 6 4 x 0 2 4 6 8 g ( x) 4 8 0 2 6 Answer: The objective is to determine 𝑓 −1 (0) . A function is said to be invertible if it sends each input x-value to exactly one output f(x). If a function f(x) is invertible, its inverse is written 𝑓 −1 (𝑥) . The inverse 𝑓 −1 (𝑥) takes output values of f(x) and produces input values. That is: 𝑓 −1 (𝑏) = 𝑎 if and only if 𝑓(𝑎) = 𝑏 . Let 𝑓 −1 (0) = 𝑎 . This means a is the number such that 𝑓(𝑎) = 0 . To find a, look for the column in the table with an f(x)-value of 0. In the given table, 𝑓(𝑎) = 0 when 𝑎 = 4 . Thus, 𝑓 −1 (0) = 4 Conclusion: Using the given table, the value of 𝑓 −1 (0) was determined to be 4. Question #3 3 Planetary orbits. The formula 𝑇(𝑥) = 𝑥 2 calculates the time in years that it takes a planet to orbit the sun if the planet is x times farther from the sun than earth is. a) Find the inverse of T b) What does the inverse of T calculate? Given: The time in years that it takes a planet to orbit the sun if the planet is x times farther from the sun than 3 earth is given by 𝑇(𝑥) = 𝑥 2 Answer: Formula used: 3 The formula 𝑇 = 𝑓(𝑥) = 𝑥 2 which converts planet orbital radius to orbital period. If 𝑇 = 𝑓(𝑥) and f(x) is a bijective mapping (both one-to-one and surjective) then its inverse exists and thus 𝑥 = 𝑓 −1 (𝑇). Calculation: 3 a) Consider the given formula 𝑇 = 𝑓(𝑥) = 𝑥 2 with 𝑥 ≥ 0 , which is an increasing function satisfying 2 bijectivity. Now, raising both the sides to power 3 2 2 3 3 2 we get 𝑇 3 = (𝑥 2 ) . This gives, 𝑥 = 𝑓 −1 (𝑇) = 𝑇 3 2 b) From part (a) we have seen 𝑥 = 𝑓 −1 (𝑇) = 𝑇 3 . Thus inverse of T calculates orbit radius as a function of orbital period. Conclusion: 2 Thus, we found out that 𝑥 = 𝑓 −1 (𝑇) = 𝑇 3 is the inverse of T and it calculates orbit radius as a function of orbital period. We can also explain it in another way. If planet orbital period function is represented by 3 2 (𝑥) = 𝑥 2 , its inverse can be expressed as 𝑇 −1 (𝑥) = 𝑥 3 . For T(x), the argument x represents the number of times the planet with orbital period T is farther from sun than Earth is. Question #4 To describe: The difference between (𝑓𝑔)(𝑥) and (𝑓 ⋅ 𝑔)(𝑥) Answer: Formula used: Product of two functions: (𝑓𝑔)(𝑥) = 𝑓(𝑥)𝑔(𝑥) Composition of two functions: (𝑓 ⋅ 𝑔)(𝑥) = 𝑓(𝑔(𝑥)) Calculation: The function (fg)(x) denotes the product of two functions f and g and can be written as (𝑓𝑔)(𝑥) = 𝑓(𝑥)𝑔(𝑥). The function (𝑓 ⋅ 𝑔)(𝑥) denotes the composition of two functions f and g and can be written as (𝑓 ⋅ 𝑔)(𝑥) = 𝑓(𝑔(𝑥)) . Let us see their examples. Suppose two functions are: 𝑓(𝑥) = 3𝑥 2 , 𝑔(𝑥) = 𝑥 + 1 Therefore, (𝑓𝑔)(𝑥) = 𝑓(𝑥)𝑔(𝑥) = 3𝑥 2 (𝑥 + 1) = 3𝑥 3 + 3𝑥 2 2 (𝑓 ⋅ 𝑔)(𝑥) = 𝑓(𝑔(𝑥)) = 3(𝑔(𝑥)) = 3(𝑥 + 1)2 = 3(𝑥 2 + 1 + 2𝑥) = 3𝑥 2 + 6𝑥 + 3 Conclusion: The function (𝑓𝑔)(𝑥) denotes the product of two functions f and g and can be written as (𝑓𝑔)(𝑥) = 𝑓(𝑥)𝑔(𝑥) . The function (𝑓 ⋅ 𝑔)(𝑥) denotes the composition of two functions f and g and can be written as (𝑓 ⋅ 𝑔)(𝑥) = 𝑓(𝑔(𝑥)). Question #5 2 The value of expression −2(27)3 . Answer: Formula used: (𝑎𝑚 )𝑛 = 𝑎𝑚𝑛 Calculation: 2 The expression is −2(27)3 Rewrite the expression as, 2 2 −2(27)3 = −2(33 )3 Now, simplify the expression. 2 2 −2(33 )3 = −2(3)3×3 = −2(3)2 = −2(9) = −18 [∵ (𝑎𝑚 )𝑛 = 𝑎𝑚𝑛 ] 2 Hence, the simplification of −2(27)3 is -18 Conclusion: With the help of formulas, we could simplify the expression. Question #6 3 0 The value of expression 5 ( ) . 4 Key Concepts Used: Any number raised to an exponent 0 is equal to 1. That is. 𝑎0 = 1 . Answer: Calculation: The power zero of any number is 1. 3 0 So, ( ) = 1 4 Now, rewrite the expression. 3 0 5 ( ) = 5(1) 4 =5 3 0 Hence, the simplification of 5 ( ) is 5. 4 Conclusion: With the help of formulas, we could simplify the given expression. Question #7 1 The value of expression (5101 )101 . Answer: Formula used: (𝑎𝑚 )𝑛 = 𝑎𝑚𝑛 Calculation: 1 The expression is (5101 )101 Rewrite the expression as, 1 1 (5101 )101 = (5)101×101 [∵ (𝑎𝑚 )𝑛 = 𝑎𝑚𝑛 ] Now, simplify the expression. 1 (5)101×101 = (5)1 =5 1 Hence, the simplification of (5101 )101 is 5. Conclusion: With the help of formulas, we could simplify the given expression. Question #8 Whether the given data is linear or exponential and find the corresponding model. Given: x 0 1 2 3 4 y 2 8 32 128 512 Answer: Formula used: 𝑦 = 𝑐𝑎𝑥 Calculation: In the given table, the data do not change by a fixed amount foreach unit increase in x. The data is not linear. Now, find the data are exponential, calculate ratios of consecutive y's. 512128 = 4,12832 = 4,328 = 4,82 = 4 For each unit increase in x, the y's multiplied by 4. So, the function is exponential. In an exponential function 𝑓(𝑥) = 𝑐𝑎𝑥 . Where 𝑐 = 𝑓(0) = 2 and 𝑎 = 4 . Substitute 2 for c and 4 for a. 𝑓(𝑥) = 𝑐𝑎𝑥 = 2(4)𝑥 Hence, the function is 𝑓(𝑥) = 2(4)𝑥 . Conclusion: On analyzing the given table, we figured out that the data fits an exponential model. The required model is 𝑓(𝑥) = 2(4)𝑥 . Question #9 The approximate final value of the amount $900 at 0.99% compounded continuously for 9 years by using the compound interest formula. The formula for total amount compounded continuously is given by 𝐴 = 𝑃𝑒 𝑟⋅𝑡 Where, A is the total amount, P is the principal, r is the annual interest rate, t is the time in years and e is the mathematical constant with approximate value 2.71828. By substituting the values in the formula, we get the total value of this given amount. Answer: Formula used: The formula for total amount compounded continuously is given by 𝐴 = 𝑃𝑒 𝑟⋅𝑡 Where, A is the total amount, P is the principal, r is the annual interest rate, t is the time in years and e is the mathematical constant with approximate value 2.71828. Calculation: Step 1: Substitute 𝑃 = 900, 𝑟 = 0.0099 , and 𝑡 = 9 in the formula to get A. 𝐴 = 𝑃𝑒 𝑟⋅𝑡 = 900𝑒 0.0099×9 = 900 × 𝑒 0.0891 = 900 × 1.0932 = $983.87 Conclusion: Using the compound interest formula, the final value of the amount $900 at 0.99% compounded continuously for 9 years was found to be $983.87. Question #10 The value of expression 6log6 (𝑥+1) Answer: Formula used: The property of logarithm 𝑏 log𝑏 𝑥 = 𝑥 Calculation: Using the rule 𝑏 log𝑏 𝑥 = 𝑥 . Here, 𝑏 = 6 and x=x+1 By applying the above rule, 𝑏 log𝑏 𝑥 = 6log6 (𝑥+1) 𝑥 =𝑥+1 Thus, the value of 𝑥 = 𝑥 + 1 Conclusion: The value of the given expression 6log6 (𝑥+1) is (𝑥 + 1) Question #11 The value of expression log 8 64 Answer: Formula: 𝑚 The property of logarithm log 𝑎𝑛 𝑎𝑚 = 𝑛 Calculation: To find the value of the given expression we need to use the rules of logarithmic function. In this problem 𝑚 we need to use the rule log 𝑎𝑛 𝑎𝑚 = 𝑛 Using the below rule. 𝑚 log 𝑎𝑛 𝑎𝑚 = 𝑛 log 8 64 = log 81 82 Here, 𝑛 = 1 and 𝑚 = 2 By applying the above rule, 𝑚 log 𝑎𝑛 𝑎𝑚 = 𝑛 log 8 64 = log 81 82 Here, 𝑛 = 1 and 𝑚 = 2 log 𝑎𝑛 𝑎𝑚 = log 8 64 log 8 64 = log 81 82 𝑚 𝑛 = 2 1 Thus, the value of 𝑚 𝑛 2 = . 1 Conclusion: The value of the given expression log 8 64 is 2. Question #12 The solution for the equation. Use the change of base formula to approximate axact answer to the nearest hundredth when approximate. 2 × 10𝑥 = 66. Answer: Formula used: Change of base formula for logarithm is given by log 𝑎 𝑥 = log𝑏 𝑥 log𝑏 𝑎 , for 𝑎 > 0 and 𝑥 > 0 Calculation: 2 × 10𝑥 = 66 66 10𝑥 = 2 10𝑥 = 33 𝑥 = log10 (33) Change to an logarithmic equation 𝑥= 𝑥= log(33) log(10) 1.5185 1 Using the Change of Base Formula log(13) = 1.5185 and log(10) = 1 𝑥 ≈ 1.52 Conclusion: The solution for the given equation is 𝑥 ≈ 1.52 . Question #13 Show that the series is convergent. How many terms of the series do we need to add in order to find the sum to the indicated accuracy? summation n=1 to infinity (−1)𝑛 + 1/𝑛6 (|𝑒𝑟𝑟𝑜𝑟| < 0.00005) Answer: Step 1 Note that, ∑∞ 𝑛=1 Clearly 𝑎𝑛 = 1 𝑛6 (−1)𝑛+1 𝑛6 is a a alternating series. is decreasing for all 𝑛 > 1 Therefore by alternating Series Test, the given series converges. Also Remember that? the error of the nth partial sum is bounded by |𝑎𝑛+1 | Therefore? we find the smallest n such that, 𝑎𝑛 < 0.00005 And we will conclude that, the sum jf first n-1 therms of the series approximate the sun within the allotted error. The summation starts from 𝑛 = 1, therefore there are n-1 terms before 𝑎𝑛