Questions and Answers #5 Function

Questions and Answers Sheet 5 Function Questions #1 Which of the following functions f has a removable discontinuity at a? If the discontinuity is removable, find a function g that agrees with f for 𝑥 ≠ 𝑎 and is continuous at a. 𝑓(𝑥) = 𝑥 4 −1 𝑥−1 ,𝑎 = 1 Answer: Given the function 𝑓(𝑥) = 𝑥 4 −1 𝑥−1 we want to see whether it has a removable discontinuity. Observe that the function f is not defined at 𝑥 = 1 and hence function is not continuous at 𝑥 = 1. Now in order se whether it is removable or not, let us find out the limit. lim 𝑓 (𝑥) = lim 𝑥 4 −1 𝑥→1 𝑥−1 𝑥→1 (𝑥 2 − 1)(𝑥 2 + 1) 𝑥→1 𝑥−1 = lim = lim (𝑥−1)(𝑥+1)(𝑥 2 +1) 𝑥−1 𝑥→1 = lim(𝑥 2 + 1) (𝑥 + 1) 𝑥→1 = lim(𝑥 2 + 1) ⋅ lim(𝑥 + 1) 𝑥→1 𝑥→1 = (1 + 1)(1 + 1) =4 Since the limit exists at 𝑥 = 1 and therefore 𝑥 = 1 is a removable discontinuity. Define the function  f ( x) if x  1 g ( x) =  if x = 1 4 Then the function g is continuous at 1 as lim g ( x) = lim f ( x) = 4 = g (1) x →1 x →1 Questions #2 Find the indicated difference quotient and simplify your answer. 𝑓(𝑥) = 𝑥 3 + 3𝑥 𝑓(𝑥) = 𝑓(𝑡) = 1 𝑥2 𝑔(𝑥)−𝑔(3) , 1 𝑡−2 𝑓(𝑥+ℎ)−𝑓(𝑥) 𝑥−3 , 𝑓(𝑡)−𝑓(1) 𝑡−1 Answer: ℎ , h does not equal 0 , x does not equal 3. , t does not equal 1. Using given 𝑓(𝑥) first find 𝑓(𝑥 + ℎ). 𝑓(𝑥) = 𝑥 3 + 3𝑥 𝑓(𝑥 + ℎ) = (𝑥 + ℎ)3 + 3(𝑥 + ℎ) 𝑓(𝑥 + ℎ) = 𝑥 3 + 3𝑥 2 ℎ + 3𝑥ℎ2 + ℎ3 + 3𝑥 + 3ℎ Plugged the expression of 𝑓(𝑥) and 𝑓(𝑥 + ℎ) in the difference quotient and simplified. 𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ = = = (𝑥 3 +3𝑥 2 ℎ+3𝑥ℎ2 +ℎ3 +3𝑥+3ℎ)−(𝑥 3 +3𝑥) ℎ 3𝑥 2 ℎ+3𝑥ℎ2 +ℎ3 +3ℎ ℎ ℎ(3𝑥 2 )+3𝑥ℎ+ℎ2 +3 ℎ 2 = 3𝑥 + 3𝑥ℎ + ℎ2 + 3 Answer: = 3𝑥 2 + 3𝑥ℎ + ℎ2 + 3 Questions #3 Find the gradient vector field of 𝑓(𝑥, 𝑦) = 𝑥𝑒 3𝑥𝑦 . Answer: Given function, 𝑓(𝑥, 𝑦) = 𝑥𝑒 3𝑥𝑦 Gradient vector field is given by, ∇𝑓 = 𝑑𝑓 𝑑𝑓 𝑖× 𝑗 𝑑𝑥 𝑑𝑦 ∇𝑓 = 𝑑 𝑑 3𝑥𝑦 3𝑥𝑦 (𝑥𝑒 )𝑖 + (𝑥 )𝑗 𝑑𝑥 𝑑𝑦 𝑒 Using product rule ∇𝑓 = 𝑥 3𝑥𝑦 ∇𝑓 = 𝑥𝑒 𝑑 𝑑𝑥 (𝑒 3𝑥𝑦 )𝑖 + 𝑒 3𝑥𝑦 𝑑 𝑑𝑥 (𝑥)𝑖 + 𝑥 𝑑 𝑑𝑦 (𝑒 {3𝑥𝑦) }𝑗 𝑑 𝑑 (3𝑥𝑦)𝑖 + 𝑒 3𝑥𝑦 𝑖 + 𝑥𝑒 3𝑥𝑦 (3𝑥𝑦)𝑗 𝑑𝑥 𝑑𝑦 ∇𝑓 = (3𝑥 + 1)𝑒 3𝑥𝑦 𝑖 + 3𝑥 2 𝑒 3𝑥𝑦 𝑗 Questions #4 For what value of the constant c is the function f continuous on (−∞, ∞) ? cx 2 + 4 x if x  5 f ( x) =  3  x − cx if x  5 Answer: The function f is defined by: The definition of the function changes at 𝑥 = 5 So the function is said to be continuous if the left hand limit and the right hand limit of the function are equal and same as the value of the function at the point. cx 2 + 4 x if x  5 f ( x) =  3  x − cx if x  5 The left hand limit of the function is given by: The right hand limit of the function is given by: And the value of the function is same as the right hand limit of the function. lim f ( x) = lim− (cx 2 + 4 x) \lim_{x\to5^-}f(x)=c(5)^2+4(5) x → 5− x →5 lim f ( x) = 25c + 20 x →5− lim+ f ( x) = lim+ ( x3 − cx) x →5 x →5 lim f ( x) = 125 − 5c x →5+ Hence, for the function to be continuous the limits must be equal: Hence the function is continuous when 𝑐 = 3.5 25𝑐 + 20 = 125 − 5𝑐 25𝑐 + 5𝑐 = 125 − 20 30𝑐 = 105 𝑐= 105 30 𝑐 = 3.5 Answer: The function f is continuous when for 𝑐 = 3.5 Questions #5 Using a directrix of 𝑦 = −2 and a focus of (2, 6), what quadratic function is created? 𝑓(𝑥) = −(𝑥 − 2)2 − 2 𝑓(𝑥) = (𝑥 − 2)2 + 2 𝑓(𝑥) = (𝑥 − 2)2 − 2 of 𝑓(𝑥) = −(𝑥 + 2)2 − 2 Answer: To find equation of quadratic function Given, directr, 𝑦 = −2 And focus = (2,6) We know, from any point (𝑥, 𝑦) on the parabola the focus and directrix are equidistant 1. Distance of (𝑥, 𝑦) from (2,6) =distance of directrix, 𝑦 = −2 from(𝑥, 𝑦) ((𝑥 − 2)2 + (𝑦 − 6)2 )0.5 = |𝑦 − (−2)| {using distance formula, if (𝑥1 , 𝑦1 )&(𝑥2 , 𝑦2 ) are two points then dist between them is ((𝑥1 − 𝑥2 )2 + (𝑦1 − 𝑦2 )2 )0.5 } ((𝑥 − 2)2 + (𝑦 − 6)2 )0.5 = |𝑦 + 2| Squaring both sides, we get (((𝑥 − 2)2 + (𝑦 − 6)2 )0.5 )2 = (|𝑦 + 2|)2 (𝑥 − 2)2 + (𝑦 − 6)2 = (𝑦 + 2)2 (𝑦 − 6)2 − (𝑦 + 2)2 = −(𝑥 − 2)2 𝑌 2 + 36– 12𝑦– (𝑦 2 + 4 + 4𝑦) = −(𝑥 − 2)2 𝑌 2 + 36– 12𝑦 − 𝑦 2 − 4– 4𝑦 = −(𝑥 − 2)2 −16𝑦 + 32 = −(𝑥 − 2)2 (𝑥 − 2)2 + 32 = 16𝑦 16𝑦 = (𝑥 − 2)2 + 32 𝑌= 1 32 (𝑥 − 2)2 + 16 16 𝑌= 1 (𝑥 − 2)2 + 2 16 Hence, the quadratic function is 𝐹(𝑥) = 1 16 (𝑥 − 2)2 + 2 Questions #6 2 Find the absolute extrema of the function on the closed interval. 𝑌 = 3𝑥 3 − 2𝑥, [−1,1] Minimum (𝑥, 𝑦) =? Maximum (𝑥, 𝑦) =? Answer: 𝑦= 3𝑥 2 − 2𝑥𝑥 ∈ [−1,1] 3 = 𝑥 2 − 2𝑥 ⇒ 𝑑𝑦 𝑑𝑥 = 2𝑥 − 2 For 𝑥 ∈ [−1,0] → 𝑥 ∈ [0,1] → 𝑑𝑦 𝑑𝑥 ≤0 𝑑𝑦 ≤0 𝑑𝑥 y is decreasing in [−1,1] ⇒ 𝑦𝑚𝑎𝑥 = 𝑦(−1) = (−1)2 − 2(−1) = 1+2= 3 𝑦(1) = 𝑦𝑚𝑖𝑛 = 1 − 2(1) 𝑦𝑚𝑎𝑥 = 3, 𝑦𝑚𝑖𝑛 = −1 Questions #7 1 Given a cost function 𝐶(𝑥) as 𝐶(𝑥) = 5,000 + 20𝑥 − 𝑥 2 , 4 calculate the marginal average when 𝑥 = 20, use your result to estimate the total cost when 𝑥 = 21, and calculate the percentage error between this estimate and the actual cost when 𝑥 = 21. Give your answer as a percentage to two decimal places. Do not include the percent symbol in your answer. Answer: 1 Given function is 𝐶(𝑥) = 5.00 + 20𝑥 − 𝑥 2 4 Then the average cost is given by So, 𝐶(𝑥) = = 𝐶(𝑥) = 𝑥 𝐶(𝑥) 𝑥 1 5.000+20𝑥− 𝑥 2 4 𝑥 5000 1 + 20 − 𝑥 𝑥 4 And the marginal average cost is given by the derivative of average cost function. So,𝐶 ′ (𝑥) = 5000 1 +20− 𝑥) 𝑥 4 𝑑( 𝑑𝑥 =− 5000 1 +0− 2 𝑥 4 =− 5000 1 − 𝑥2 4 So, 𝐶 ′ (20) = − And 𝐶 ′ (20) = 5000 202 5000 20 1 − = −12.5 − 0.25−= 12.75 4 1 + 20 − (20) = 265 4 We can write 𝐶 ′ (21)as given below using the derivative 𝐶 ′ (21) = 𝐶 ′ (20) + 𝐶 ′ (20) = 265 − 12.75 = 252.25 𝐶(21) = 252.25 21 𝐶(21) = 5297.25 And exact cost when 𝑥 = 21 1 𝐶𝐸 (21) = 5,000 + 20(21) − (21)2 4 = 5309.75 So, percentage error between the estimated and actual cost is So, when 𝑥 = 21 then percentage error is 5309.75−5297.25 5309.75 𝐶𝐸 (𝑥)−𝐶(𝑥) × 100 = 0.24% Questions #8 Find the absolute extrema of the function on the closed interval. 2 𝑦 = 3𝑥 3 − 2𝑥, [−1,1] Answer: 2 𝑦 = 3𝑥 3 − 2𝑥, [−1,1] So, Critical number: 𝐶𝐸 × 100 2 𝑦 = 3𝑥 3 − 2𝑥 diffirentiate to x 2 1 𝑦 ′ = 3 𝑥 −3 − 2 𝑥 1 𝑦 ′ = 2𝑥 −3 − 2 𝑦′ = 2 1 −2 𝑥3 Put 𝑦 ′ = 0 ⇒ 2 1 −2=0 𝑥3 ⇒ 2 1 =2 𝑥3 ⇒ 1 1 =1 𝑥3 1 ⇒ 1 = 𝑥3 ⇒ 13 = 𝑥 ⇒1=𝑥 ⇒𝑥=1 Considering boundary numbers and critical numbers, we have the following numbers: 𝑥 = −1 and 𝑥 = 1 2 𝑓(𝑥) = 𝑦 = 3𝑥 3 − 2𝑥 𝑓(−1) = 3(−1)2/3 − 2(−1) = 3(1) + 2 =5 2 𝑓(1) = 3(1)3 − 2(1) = 3−2 =1 f ( −1) = 5  absolute maximum f (1) = 1  absolute minimum Questions #9 Apply Newton’s Method using the given initial guess, and explain why the method fails. 𝑦 = 𝑥 3 − 2𝑥 − 2, 𝑥1 = 0 Answer: 𝑦 = 𝑥 3 − 2𝑥 − 2 with an initial guess of 𝑥1 = 0 By Newton's method, the next approximation of the root can be found as: 𝑥𝑛+1 = 𝑥𝑛 − 𝑑(𝑥𝑛 ) 𝑓 ′ (𝑥𝑛 ) Let 𝑓(𝑥) = 𝑥 3 − 2𝑥 − 2 𝑓 ′ (𝑥) = 3𝑥 2 − 2 Therefore, 𝑥𝑛+1 = 𝑥𝑛 − 𝑥𝑛3 − 2𝑥𝑛 − 2 3𝑥 2 − 2 Calculate next approximation 𝑥2 : 𝑥2 = 0 − =− 0 − 2(0) − 2 3(0) − 2 −2 −2 = −1 Calculate 𝑥3 . 𝑥3 = −1 − = −1 − (−1)3 − 2(−1) − 2 3(−1)2 − 2 −1 1 = −1 + 1 =0 Similarly, the next approximiation be -1. Questions #10 Find the Taylor polynomial 𝑇3 (𝑥) for the function f centerd at the number a. 𝑓(𝑥) = 𝑒 −4𝑥 sin(2𝑥) , 𝑎 = 0 Answer: The function is 𝑓(𝑥) = 𝑒 −4𝑥 sin(2𝑥) The formula to find the Taylor polynomila for the function f centered at the number a, is 𝑛 𝑇𝑛 (𝑥) = ∑ 𝑖=0 𝑓 (𝑖) (𝑎) (𝑥 − 𝑎)𝑡 𝑖! For 𝑛 = 3 3 𝑇3 (𝑥) = ∑ 𝑖=0 𝑓 (𝑖) (𝑎) (𝑥 − 𝑎)𝑡 𝑖! = 𝑓(𝑎) + 𝑓 ′ (𝑎)(𝑥 − 𝑎) + 𝑓 ′′ (𝑎) 𝑓 ′′′ (𝑎) (𝑥 − 𝑎)2 + (𝑥 − 𝑎)3 2! 3! Obtained the first, second and third derivatives as follows. 𝑓 ′ (𝑥) = 2𝑒 −4𝑥 cos(2𝑥) − 4𝑒 −4𝑥 sin(2𝑥) 𝑓 ′′ (𝑥) = 12𝑒 −4𝑥 sin(2𝑥) − 16𝑒 −4𝑥 cos(2𝑥) 𝑓 ′′′ (𝑥) = 88𝑒 −4𝑥 cos(2𝑥) − 16𝑒 −4𝑥 sin(2𝑥) At 𝑎 = 0, the function and the derivatives are 𝑓(0) = 𝑒 −4(0) sin(2(0)) = 0 𝑓 ′ (0) = 2𝑒 −4(0) cos(2(0)) − 4𝑒 −4(0) sin(2(0)) = 2 𝑓 ′′ (0) = 12𝑒 −4(0) sin(2(0)) − 16𝑒 −4(0) cos(2(0)) = −16 𝑓 ′′′ (0) = 88𝑒 −4(0) cos(2(0)) − 16𝑒 −4(0) sin(2(0)) = 88 Thus, the Taylor polynomial becomes 𝑇3 (𝑥) = 𝑓(0) + 𝑓 ′ (0)(𝑥 − 0) + = 0 + 2𝑥 + 𝑓 ′′ (0) 𝑓 ′′ (0) 𝑓 ′′′ (0) (𝑥 − 0)2 + (𝑥 − 0)2 + (𝑥 − 0)3 2! 2! 3! −16 2 88 3 𝑥 + 𝑥 2 6 = 2𝑥 − 8𝑥 2 + 44 3 𝑥 3 Therefore, the Taylor polynomial for the function f centered at the number a, 𝑇3 (𝑥) = 2𝑥 − 8𝑥 2 + 44 3 𝑥 3 Questions #11 Determine if the graph is symmetric about the x-axis, the y-axis, or the origin. Answer: r = 9sin 7 Replace (𝑟, 𝜃) by (𝑟, −𝜃) , we get ⇒ 𝑟 = 9 sin 7 (−𝜃) ⇒ 𝑟 = −9 sin 7 𝜃. Which is not same as original equation. Therefore given equation is not symmetric with respect to x-axis. Replace (𝑟, 𝜃) by (−𝑟, −𝜃) , we get ⇒= 9 sin 7 (−𝜃) ⇒ 𝑟 = 9 sin 7 𝜃. Which is same as original equation. Therefore given equation is symmetric with respect to y-axis. Questions #12 Find f. 𝑓 ′ (𝑡) = 20 1 + 𝑡2 ′ , 𝑓(1) = 0 Answer: Given, function is r = 9sin 7 𝑓 ′ (𝑡) = 20 , 𝑓(1) = 0 1 + 𝑡2 Integrate with respect to ′𝑡 ′ 𝑓(𝑡) = ∫ 𝑓 ′ (𝑡) + 𝐶 =∫ 20 +𝐶 1 + 𝑡2 Apply the identity, ∫ 𝑎2 1 1 𝑥 = tan−1 ( ) 2 +𝑡 𝑎 𝑎 Hence, ⇒ 𝑓(𝑡) = 20 tan−1 (𝑡) + 𝐶 Apply initial condition, ⇒ 𝑓(1) = 20 tan−1 (1) + 𝐶   0 = 20 tan −1 (tan ) + C 4 ⇒ 0 = 20 ⋅ 𝜋 +𝐶 4 ⇒ 𝐶 = −5𝜋 Questions #13 Find a potential function f for the field F. 𝐹 = 2𝑥𝑖 + 3𝑦𝑗 + 4𝑧𝑘 Answer: If 𝐹 = ∇𝑓 for some scalar function f, then 𝑀 = 2𝑥 = 𝑑𝑓 𝑑𝑥 𝑁 = 3𝑦 = 𝑑𝑓 𝑑𝑦 𝑃 = 4𝑧 = 𝑑𝑓 𝑑𝑧 From equation we have that 𝑑𝑓 𝑑𝑥 = 2𝑥 ⇒ 𝑓 = ∫ 2𝑥𝑑𝑥 = 𝑥 2 + 𝐶(𝑦, 𝑧) where C(y,z) does not depended on x. Therefore 𝑓(𝑥, 𝑦, 𝑧) = 𝑥 2 + 𝐶(𝑦, 𝑧) Substituting this result into equation we have that 𝑑𝑓 𝑑 2 [𝑥 + 𝐶(𝑦, 𝑧)] = 3𝑦 = 3𝑦 ⇒ 𝑑𝑦 𝑑𝑦 ⇒ 𝑑𝐶 𝑦2 = 3𝑦 ⇒ 𝐶 = 3 + 𝐶0 (𝑧) 𝑑𝑦 2 Where 𝐶0 is a function of z. Therefore, 3 𝑓(𝑥, 𝑦, 𝑧) = 𝑥 2 + 𝑦 2 + 𝐶0 (𝑧) 2 Finally, by substituting this result into equation we have that 𝑑𝑓 𝑑 3 = 4𝑧 ⇒ (𝑥 2 + 𝑦 2 + 𝐶0 (𝑧)) = 4𝑧 𝑑𝑧 𝑑𝑧 2 ⇒ 𝑑𝐶0 = 4𝑧 ⇒ 𝐶0 ∫ 4𝑧𝑑𝑧 𝑑𝑧 ⇒ 𝐶0 = 2𝑧 2 + 𝑐 3 where c is a real constant. Letting 𝑐 = 0 we have that 𝑓(𝑥, 𝑦, 𝑧) = 𝑥 2 + 𝑦 2 + 2𝑧 2 is a potential function 2 for F. Questions #14 Verify that the following functions are solutions to the given differential equation. 𝑦 = 2𝑒 𝑥 − 𝑥 − 1 solves 𝑦′ = 𝑦 + 𝑥 Answer: 𝑦 = 2𝑒 𝑥 − 𝑥 − 1 solves 𝑦′ = 𝑦 + 𝑥 𝑦 = 2𝑒 𝑥 − 𝑥 − 1 ⇒ 𝑦′ = 𝑑 (2𝑒 𝑥 − 𝑥 − 1) 𝑑𝑥 = 2𝑒 𝑥 − 1 + 0 = (2𝑒 𝑥 − 𝑥 − 1) + 𝑥 =𝑦+𝑥 𝑦′ = 𝑑 (2𝑒 𝑥 − 𝑥 − 1) = 2𝑒 𝑥 − 1 + 0 𝑑𝑥 = (2𝑒 𝑥 − 𝑥 − 1) + 𝑥 = 𝑦 + 𝑥 Questions #15 I was just wondering what the derivative of 𝑓(𝑥) = 𝑚𝑎𝑥(0,1 − 𝑥)2 would be. What technique do you use to determine this derivative? Answer: It might be of help to sketch the function or write it without the max. We get (1 − x) 2 f ( x) =  0 if x  1 if x  1 It is easy to work out the derivative everywhere except at 𝑥 = 1. At 𝑥 = 1, work out explicitly from definition. lim+ 𝑓(1 + ℎ) − 𝑓(1) =0 ℎ lim− 𝑓(1 + ℎ) − 𝑓(1) ℎ2 = lim− = 0 ℎ→0 ℎ ℎ ℎ→0 ℎ→0 Hence, we have 2( x − 1) if x  1 f ( x) =  if x  1 0 Questions #16 Find the Taylor polynomials of degree n approximating the functions for x near 0. (Assume p is a constant) tan 𝑥 , 𝑛 = 3,4 Answer: Function 𝑓(𝑥) is given as 𝑓(𝑥) = tan 𝑥 And Now we need to calculate the derivatives of the above function at 𝑥 = 0 𝑓 ′ (𝑥) = 𝑠𝑒𝑐 2 𝑥 𝑓 ′ (0) = 𝑠𝑒𝑐 2 0 = 1 𝑓′′(𝑧𝑥 = 2 𝑡𝑎𝑛 𝑥 𝑠𝑒𝑐 ^{2} 𝑥 𝑓 ′′ (0) = 2 tan 0 sec 2 0=0 𝑓 (3) (𝑥) = 4 tan2 𝑥 sec 2 𝑥 + 2 sec 4 𝑥 𝑓 (3) (0) = 4 tan2 0 sec 2 0 + 2 sec 4 0 = 2 𝑓 (4) (𝑥) = 8 tan3 𝑥 sec 2 𝑥 + 16 tan 𝑥 sec 4 𝑥 𝑓 (4) (0) = 8 tan3 0 sec 2 0 + 16 tan 0 sec 4 0 = 0 Since the Taylor Polynomial of Degree n Approximating f(a) for x near 0 is given as 𝑓(𝑥) ≈ 𝑃𝑛 (𝑥) = 𝑓(0) + 𝑓 ′ (0)(𝑥) + 𝑓 ′ (0) 𝑓 (𝑛) (0) (𝑥)2 + ⋯ + (𝑥)𝑛 2! 𝑛! We call 𝑃𝑛 (𝑥) the Taylor polynomial of degree n centered at 𝑥 = 0 or the Taylor polynomial about 𝑥 = 0 𝑛−3 𝑓(𝑥) ≈ 𝑃3 (𝑥) = 𝑓(0) + 𝑓 ′ (0)(𝑥) + = 0 = (1)𝑥 + (0) =𝑥+ 𝑓 ′ (0) 𝑓 (𝑛) (0) (𝑥)2 + ⋯ + (𝑥)𝑛 2! 𝑛! 𝑥2 𝑥3 + (2) 2! 3! 𝑥3 3 𝑛=4 𝑓(𝑥) ≈ 𝑃4 (𝑥) = 𝑓(0) + 𝑓 ′ (0)(𝑥) + = 0 + (1)𝑥 + (0) =𝑥+ 𝑛=3 𝑥3 3 𝑓 ′ (0) 𝑓 (𝑛) (0) (𝑥)2 + ⋯ + (𝑥)𝑛 2! 𝑛! 𝑥3 𝑥3 𝑥4 + (2) + (0) 2! 3! 4! 𝑓(𝑥) = 𝑥 + 𝑥3 3 𝑛=4 𝑓(𝑥) = 𝑥 + 𝑥3 3 Questions #17 If (𝑥, 𝑦) ⇒ (𝑥 + 3, 𝑦 − 7) maps 𝑇𝑈𝑉𝑊 to 𝑇`𝑈`𝑉`𝑊`, what is the algebraic description what willmap 𝑇`𝑈`𝑉`𝑊` to 𝑇𝑈𝑉𝑊 Answer: Given (𝑥, 𝑦) ⇒ (𝑥 + 3, 𝑦 − 7)𝑥, 𝑦 ⇒ 𝑥 + 3, 𝑦 − 7 maps 𝑇𝑈𝑉𝑊 to 𝑇 ′ 𝑈 ′ 𝑉 ′ 𝑊 ′ . Let (𝑥, 𝑦) ⇒ (𝑎, 𝑏)(𝑥, 𝑦) ⇒ (𝑎, 𝑏) where 𝑎 = 𝑥 + 3, 𝑏 = 𝑦 − 7𝑥 = 𝑎 − 3, 𝑦 = 𝑏 + 7 Therefore, (𝑎, 𝑏) ⇒ (𝑎 − 3, 𝑏 + 7)𝑎, 𝑏 ⇒ 𝑎 − 3, 𝑏 + 7 Therefore, (𝑥, 𝑦) ⇒ (𝑥 − 3, 𝑦 + 7)𝑥, 𝑦 ⇒ 𝑥 − 3, 𝑦 + 7 maps 𝑇 ′ 𝑈 ′ 𝑉 ′ 𝑊 ′ to 𝑇𝑈𝑉𝑊 Finally answer : (𝑥, 𝑦) ⇒ (𝑥 − 3, 𝑦 + 7)𝑥, 𝑦 ⇒ 𝑥 − 3, 𝑦 + 7 maps 𝑇 ′ 𝑈 ′ 𝑉 ′ 𝑊 ′ to 𝑇𝑈𝑉𝑊 Questions #18 Find equations of both lines through the point (2, −3) thatare tangent to the parabola 𝑦 = 𝑥 2 + 𝑥 Answer: Given: The given function is = 𝑥 2 + 𝑥 Find the derivative: = 𝑦 = 𝑥2 + 𝑥 = 𝑑𝑦 = 2𝑥 + 1 𝑑𝑥 Find the equation of the tangent line: Since 𝑦 = 𝑥 2 + 𝑥 . Considerthe point on the porabola (𝑎, 𝑎2 + 𝑎) Thus the slope of the line at (𝑎, 𝑎2 + 𝑎) is 𝑚 = 𝑑𝑦 𝑑𝑥 = 2𝑎 + 1 By using the slope intercept form = 𝑦 = 𝑚𝑥 + 𝑐 ⇒ 𝑦 = (2𝑎 + 1)𝑥 = 𝑐 If the line is a tangent to a given parabola,then it passes through (𝑎, 𝑎2 + 𝑎) = 𝑎2 + 𝑎 = (2𝑎 + 1) + 𝑐 = 𝑎2 + 𝑎 = 2𝑎2 + 1 + 𝑐 = 𝑐 = −𝑎2 Thus the equation of the langent line is = 𝑦 = (2𝑎 + 1)𝑥 − 𝑎2 . The required tangent passes through the point (2, −3) Substitute = 𝑥 = 2 and = 𝑦 = −3 in above equation = −3 = (2𝑎 + 1)2 − 𝑎2 = −3 = 4𝑎 + 2 − 𝑎2 = 𝑎2 − 4𝑎 − 5 = 0 = (𝑎 + 1)(𝑎 − 5) = 0 = 𝑎 = −1 or = 𝑎 = 5 Substitute = 𝑎 = −1 in = 𝑦 = (2𝑎 + 1)𝑥 − 𝑎2 = 𝑦 = (2(−1) + 1)𝑥 − (−1)^{2}) = 𝑦 = −𝑥 − 1 Substitute = 𝑎 = 5 in 𝑦 = (2𝑎 + 1)𝑥 − 𝑎2 = 𝑦 = (2(5) + 1)𝑥 − (5)2 = 𝑦 = 11𝑥 − 25 Questions #19 Find the point on the line 𝑦 = 4𝑥 + 5 that is closest to the origin. (𝑥, 𝑦) = Answer: Step 1 Find the point on the line 𝑦 = 4𝑥 + 5 that is closest to the origin. Solution: let's say (ℎ, 𝑘) is the closest to the origin (0,0) and lies on the line 𝑦 = 4𝑥 + 5 . So (ℎ, 𝑘) will satisfy the line 𝑦 = 4𝑥 + 5 equation Put 𝑥 = ℎ , 𝑦 = 𝑘 , 𝑘 = 4ℎ + 5 (1) distance of point (ℎ, 𝑘) from (0,0). 𝑑 = √(ℎ − 0)2 + (𝑘 − 0)2 𝑑 = √{ℎ2 + 𝑘 2 (2) NST Shortest distance of the given line from origin [4𝑋(0)(0) + 5] √42 + (−1)2 Put 4𝑥 + 5 − 𝑦 = 0 Put 𝑥 = 0, 𝑦 = 0 𝑑= 5 √17 (3) from equation (2) & (3) , √ℎ 2 + 𝑘 2 = 5 √17 ℎ2 + 𝑘 2 = 25 17 (4) Step 2 Put 𝑘 = 4ℎ + 5 into equation (4) ℎ2 + (4ℎ + 5)2 = 25 17 ℎ2 + 162 + 25 + 40ℎ = 17ℎ2 + 40ℎ + 25 − 25 17 25 =0 17 289ℎ2 + 680ℎ + 425 − 25 =0 17 289ℎ2 + 680ℎ + 400 = 0 Use sridharacharya method to find root, 𝐴 = 289, 𝐵 = 680, 𝐶 = 400 ℎ= −𝐵 ± √𝐵2 − 4𝐴𝐶 2𝐴 = −680 ± √(680)2 4 × 289 × 400 2 × 289 = −680 ± 0 2 × 289 = −40 2 × 17 = −20 17 ℎ= −20 17 so, 𝑘 = 4ℎ + 5 = −80 +5 17 = 5 17 NST 𝑘= 5 17 Hence required points are ( h, k ) = ( −20 5 , ) 17 17 Questions #20 The cost function for production of a commodity is 𝐶(𝑥) = 339 + 25𝑥 − 0.09𝑥 2 + 0.0004𝑥 3 Compare 𝐶’(100) with the cost of producing the 101st item. Answer: 𝐶(𝑥) = 339 + 25𝑥 − 0.09𝑥 2 + 0.0004𝑥 3 𝐶 ′ (𝑥) = 25 − 0.18𝑥 + 0.0012𝑥 2 (marginal cost) Increase in cost as 𝑥 = 100 units 𝐶 ′ (100) = 19 Cost of 101st item: 𝐶(101) − 𝐶(100) ≈ 2358.03 − 2339 ≈ 19.03 𝐶 ′ (100) is approximately the same as the increase in cost to produce the 101st item. Result: 𝐶 ′ (100) is approximately the same as the increase in cost to produce the 101st item.