Questions and Answers #6 Functions

Questions and Answers Sheet 6 Functions Question #1 Use the linear approximation (1 + 𝑥)𝑘 ≈ 1 + 𝑘𝑥 to find an approximation for the function f(x) for values of x near zero. 𝑓(𝑥) = (1 − 𝑥)6 Answer: Use the linear approximation (1 + 𝑥)𝑘 ≈ 1 + 𝑘𝑥 to find an approximation for the following functions for values of x near zero. 𝑓(𝑥) = (1 − 𝑥)6 = [1 + (−𝑥)]6 ≈ 1 + 6(−𝑥) = 1 − 6𝑥 Result: 𝑓(𝑥) ≈ 1 − 6𝑥 Question #2 Use the linear approximation (1 + 𝑥)𝑘 ≈ 1 + 𝑘𝑥 to find an approximation for the function f(x) for values of x near zero. 𝑓(𝑥) = 2 1−𝑥 Answer: Use the linear approximation (1 + 𝑥)𝑘 ≈ 1 + 𝑘𝑥 to find an approximation for the following functions for values of x near zero. 𝑓(𝑥) = 2 = 2[1 + (−𝑥)]−1 ≈ 2[1 + (−1)(−𝑥)] = 2 + 2𝑥 1−𝑥 Result: 𝑓(𝑥) ≈ 2 + 2𝑥 Question #3 Use the linear approximation (1 + 𝑥)𝑘 ≈ 1 + 𝑘𝑥 to find an approximation for the function f(x) for values of x near zero. 𝑓(𝑥) = 1 √1 + 𝑥 Answer: Use the linear approximation (1 + 𝑥)𝑘 ≈ 1 + 𝑘𝑥 to find an approximation for the following functions for values of x near zero. 𝑓(𝑥) = 1 1 1 𝑥 = (1 + 𝑥)−2 ≈ 1 + (− ) 𝑥 = 1 − 2 2 1 + 𝑥 √ Result: 𝑓(𝑥) ≈ 1 − 𝑥 2 Question #4 Use the linear approximation (1 + 𝑥)𝑘 ≈ 1 + 𝑘𝑥 to find an approximation for the function f(x) for values of x near zero. 𝑓(𝑥) = √2 + 𝑥 2 Answer: Use the linear approximation (1 + 𝑥)𝑘 ≈ 1 + 𝑘𝑥 to find an approximation for the following functions for values of x near zero. 1 𝑓(𝑥) = √2 + 𝑥2 𝑥2 2 1 3 𝑥2 = √2 (√1 + ) = √2 (1 + ) ≈ √2 (1 + ∗ 𝑥) = √2 (1 + ) 2 2 2 4 4 𝑥2 Result: 𝑓(𝑥) ≈ √2 (1 + 𝑥2 4 ) Question #5 Use the linear approximation (1 + 𝑥)𝑘 ≈ 1 + 𝑘𝑥 to find an approximation for the function f(x) for values of x near zero. 1 𝑓(𝑥) = (4 + 3𝑥)(3) Answer: Use the linear approximation (1 + 𝑥)𝑘 ≈ 1 + 𝑘𝑥 to find an approximation for the following functions for values of x near zero. 1 𝑓(𝑥) = (4 + 1 3𝑥)3 = 1 43 1 1 3𝑥 3 1 3 𝑥 (1 + ) ≈ 43 (1 + ∗ 𝑥) = 43 (1 + ) 4 3 4 4 Result: 1 𝑥 𝑓(𝑥) ≈ 43 (1 + ) 4 Question #6 Use the linear approximation (1 + 𝑥)𝑘 ≈ 1 + 𝑘𝑥 to find an approximation for the function f(x) for values of x near zero. 3 𝑓(𝑥) = √(1 − 𝑥 2 ) 2+𝑥 Answer: Use the linear approximation (1 + 𝑥)𝑘 ≈ 1 + 𝑘𝑥 to find an approximation for the following functions for values of x near zero. 2 2 𝑥 2 𝑥 3 𝑥 3 ) = (1 − ) = [1 + (− )] 2+𝑥 2+𝑥 2+𝑥 2 𝑥 2𝑥 2𝑥 6 + 3𝑥 − 2𝑥 6+𝑥 ≈ 1 + (− )= 1− =1− = = 3 2+𝑥 3(2 + 𝑥) 6 + 3𝑥 6 + 3𝑥 6 + 3𝑥 3 𝑓(𝑥) = √(1 − Result: 𝑓(𝑥) ≈ 6+𝑥 6+3𝑥 Question #7 Find the Wroian of the given pair of functions. cos 𝑡 , 𝑠𝑖𝑛 Answer: Let 𝑦1 = cos 𝑡 𝑎𝑛𝑑𝑦2 =. The derivatives are given on the left. 𝑦1′ = −𝑠𝑖𝑛 y2 = cos. The Wroian is given by 𝑦1 𝑦2′ − 𝑦1′ 𝑦2 . 𝑦1 𝑦2′ − 𝑦1′ 𝑦2 = (cos 𝑡)(cos 𝑡) − (− sin 𝑡)(sin 𝑡) = cos 2 t + sin 2 t = 1. Result: 1 Question #8 The inverse sine, inverse cosine, and inverse tangent functions have the following domains and ranges. The function sin−1 has domain ____ and range ____. The function cos −1 has domain ____ and range ____. The function tan−1 has domain ____ and range _____. (a) (b) (c) Answer: Solution: π π (a)The inverse sine has a domain of [-1,1], and a range of [− , ] 2 2 (b)The inverse cosine has a domain of [-1,1], and a range of [0, π] π π (c)The inverse tangent has a domain of R, and a range if (− , ) 2 2 Result: π π (a) [−1,1], [− , ] 2 2 (b) [−1,1], [0, π] π π (c) 𝑅, (− , ) 2 2 Question #9 Determine the set of points at which the function is continuous. 𝑓(𝑥, 𝑦) = ( 𝑥2𝑦3 2𝑥 2 +𝑦 2 ) if (x,y) not equal to (0,0) and 1 if (𝑥, 𝑦) = (0,0) Answer: On 𝑅𝟚 without the origin, f is a rational function, and so is continuous at every point in its domain. Since it is defined on this entire domain, we see that f is continuous at every point. To check whether f is continuous at the origin, we must check whether lim (𝑥,𝑦)⇒(0,0) 𝑓 (𝑥, 𝑦) = 𝑓(0,0). Checking along the line 𝑦 = 0, we see that lim 𝑥 2 (0)3 𝑥⇒0 2𝑥 2 +(0)2 = lim 0 = 0 𝑥⇒0 But this implies that if f has a limit as (x,y) tends to (0,0), it must be 0, which is not the value of f(0,0). Thus we see that in this case, f(x,y) is not continuous at the origin. Result: f(x,y) is not continuous at the origin. Question #10 Use a linear approximation (or differentials) to estimate the given number. √99.8 Answer: Approximation can be done by using a linearized model of our function: 𝐿(𝑥) ≈ 𝑓(𝑎) + 𝑓 ′ (𝑎) ∗ (𝑥 − 𝑎) Now we need to find the derivative of f(x): 𝑓(𝑥) = √𝑥 1 ′ 1 1 1 1 𝑓 ′ (𝑥) = (𝑥 2 ) = 𝑥 (2−1) = 𝑥 −2 2 = 2 1 2 √𝑥 Without a calculator we don't know the value of √99.8 , but since we need an approximation, we'll use the closest value that we know and that's √100 . Hence: 𝑓(𝑥) = √𝑥 = √100 = 10 𝑓 ′ (𝑥) = = 1 2√ 𝑥 = 1 2√100 = 1 2 ∗ 10 1 20 Now we can calculate the approximation using the linearized model: 𝐿(𝑥) ≈ 𝑓(𝑎) + 𝑓 ′ (𝑎) ∗ (𝑥 − 𝑎) 𝐿(99.8) = √99.8 ≈ 𝑓(100) + 𝑓 ′ (100) ∗ (99.8 − 10) ≈ 10 + 1 ∗ (44 − 45) 20 √99.8 ≈ 9.95 Result: √99.8 ≈ 9.95 Question #11 A rectangle has area 16𝑚2 . Express the perimeter of the rectangle as a function of the length of one of its sides. Answer: We must represent the sides of a rectangle by two variables. Let us chose x to be the width and y to be the length. We can now represent the area of the rectangle, knowing that the area is 16. 𝑥𝑦 = 16 𝑦= 16 𝑥 This is the perimeter of a rectangle. 𝑃 = 2𝑥 + 2𝑦 16 𝑃 = 2𝑥 + 2 ( ) 𝑥 𝑃 = 2𝑥 + 𝑃= 32 𝑥 2𝑥 2 + 32 𝑥 Note that the domain will be 𝑥 > 0 as the width must be positive. If the length is larger than the width then the domain will be 0 < 𝑥 < 4 which can be seen as follows: 16 𝑥 > 𝑥 ⇒ 16 > 𝑥 2 ⇒ 𝑥 < 4 . Result: 𝑃= 2𝑥 2 + 32 𝑥 Question #12 What is the inverse function of 𝑥 3 ? Answer: 1 1 1 𝑦 = 𝑥 3 ⇒ (𝑦)3 = (𝑥 3 )3 ⇒ 𝑦 3 = 𝑥 1 𝑦 −1 = 𝑥 3 Question #13 Simplify (−3𝑥)3 . Answer: It's easy: (−3𝑥)3 = (−3𝑥)(−3𝑥)(−3𝑥) = −27𝑥 3 Question #14 Two waves in one string are described by the wave functions: 𝑦1 = (3.0𝑐𝑚) cos(4.0 × −1.6𝑡) 𝑦2 = (4.0𝑐𝑚) sin(5.0 × −2.0𝑡) where y and x are in centimeters and t is inseconds. Find the superposition of the waves 𝑦1 + 𝑦2 at the points (a) 𝑥 = 1.00, 𝑡 = 1.00; (b)𝑥 = 1.00, 𝑡 = .500; and (c) 𝑥 = 0.500, 𝑡 = 0. Answer: Given two waves are described as 𝑦1 = (3.0𝑐𝑚) cos(4.0 × −1.6𝑡) &𝑦2 = (4.0𝑐𝑚) sin(5.0 × −2.0𝑡) a) At 𝑥 = 1𝑐𝑚 & 𝑡 = 1.0𝑠 we have 𝑦 = 𝑦1 + 𝑦2 = (3.0𝑐𝑚) cos(4.0 ∗ 1 − 1.6 ∗ 1) + (4.0𝑐𝑚) sin(5.0 ∗ 1 − 2.0 ∗ 1) = 3 cos(2.4𝑟𝑎𝑑) + 4 sin(3𝑟𝑎𝑑) = −2.212 + 0.5645 = −1.647𝑐𝑚 b) At 𝑥 = 1.00, 𝑡 = 0.500 we have 𝑦 = 𝑦1 + 𝑦2 = (3.0𝑐𝑚) cos(4.0 ∗ 1 − 1.6 ∗ 0.5) + (4.0𝑐𝑚) sin(5.0 ∗ 1 − 2.0 ∗ 0.5) = −2.995 + (−3.027) = −6.022𝑐𝑚 c) At 𝑥 = 0.500, 𝑡 = 0 we have 𝑦 = 𝑦1 + 𝑦2 = (3.0𝑐𝑚) cos(4.0 ∗ 0.5 − 1.6 ∗ 0) + (4.0𝑐𝑚) sin(5.0 ∗ 0.5 − 2.0 ∗ 0) = − 1.248 + 2.394 = 1.146𝑐𝑚 Question #15 How do you find (𝑔 − 𝑓)(𝑥) given 𝑔(𝑥) = −𝑥 2 − 1 − 2𝑥 and 𝑓(𝑥) = 𝑥 + 5 Answer: We know that (𝑔 − 𝑓)(𝑥) = 𝑔(𝑥) − 𝑓(𝑥) Explanation: (𝑔 − 𝑓)(𝑥) = 𝑔(𝑥) − 𝑓(𝑥) = −𝑥 2 − 1 − 2𝑥 − 𝑥 − 5 = −𝑥 2 − 3𝑥 − 6 so (𝑔 − 𝑓)(𝑥) = −𝑥 2 − 3𝑥 − 6 Question #16 How do you evaluate 𝑔(−5) given 𝑔(𝑥) = 3𝑥? Answer: Explanation: to evaluate 𝑔(−5) substitute 𝑥 = −5 into g(x) ⇒ 𝑔(−5) = 3 × −5 = −15 Question #17 Find the first partial derivatives of the function. 𝑧 = 𝑥 sin(𝑥𝑦) Answer: 𝑓(𝑥, 𝑦) = 𝑥 sin(𝑥𝑦) To find 𝑓𝑥 we treeat y as constant and differentiate respect to x 𝑓𝑥 = sin(𝑥𝑦) + 𝑥𝑦 cos(𝑥𝑦) To find 𝑓𝑦 we treat x as constant and differentiate with respect to y 𝑓𝑦 = 𝑥 2 cos(𝑥𝑦) Question #18 How do you find (𝑓 ⋅ 𝑔)(𝑥) and (𝑔 ⋅ 𝑓)(𝑥) and determine if the given functions are inverses of each other 𝑓(𝑥) = 𝑥 2 − 3 and 𝑔(𝑥) = √𝑥 + 3 ? Answer: To find f(g(x)) and g(f(x)) substitute, g(x) for x in f(x), and f(x) for x in g(x), respectively. If they are inverses, both substitutions will equal x. Explanation: 2 𝑓(𝑔(𝑥)) = (√𝑥 + 3) − 3 𝑓(𝑔(𝑥)) = 𝑥 + 6√𝑥 + 9 − 3 𝑓(𝑔(𝑥)) = 𝑥 + 6√𝑥 + 6 𝑔(𝑓(𝑥)) = √𝑥 2 − 3 + 3 They are not inverses, because both cases must reduce to x. Question #19 Given the following: 𝑃(𝐴) = 0.3 𝑃(𝐵) = 0.2 𝑃(𝐴 𝑎𝑛𝑑 𝐵) = 0.06 A and B are independent? Answer: Solution: Given that 𝑃(𝐴) = 0.3 𝑃(𝐵) = 0.2 𝑃(𝐴𝑎𝑛𝑑𝐵) = 0.06 𝑃(𝐴𝑎𝑛𝑑𝐵) = 𝑃(𝐴) × 𝑃(𝐵) 0.06 = 0.3 × 0.2 A nad B are independent. Question #20 Find a solution of 𝑥 𝑑𝑦 𝑑𝑥 = 𝑦 2 − 𝑦 that passes through the indicated points. a. (0,0) 1 1 b. ( , ) 4 4 1 c. (6, ) 8 Answer: By separing variables in 𝑥 𝑑𝑦 𝑑𝑥 = 𝑦 2 − 𝑦 ; we have 𝑑𝑦 𝑑𝑥 = 𝑦2 − 𝑦 𝑥 𝑑𝑦 𝑑𝑥 = 𝑦(𝑦 − 1) 𝑥 [ −1 1 𝑑𝑥 ]+[ ] 𝑑𝑦 = 𝑦 𝑦−1 𝑥 On integration, we obtain − ln(𝑦) + ln(𝑦 − 1) = ln(𝑥) + ln 𝑐 𝑦−1 ln ( ) = ln(𝑥𝑐) 𝑦 𝑦−1 = 𝑥𝑐 𝑦 𝑦 − 1 = 𝑥𝑦𝑐 𝑦 − 𝑥𝑦𝑐 𝑦(1 − 𝑥𝑐) = 1 1 1 − 𝑥𝑐 𝑦= 𝑦= 1 1−𝑐𝑥 is a solution and 𝑦 = 0 is another constant solution a) At (0,0) 𝑦= 0= 1 1−𝑥𝑐 1 1−0 0=1 This is not possible Therefore in this case there is no solution.