Questions and Answers #7 Functions

Questions and Answers Sheet 7 Function Question #1 Given the velocity 𝑣(𝑡) of an object moving along a ccordinate line at time t, find the object's displacement 𝑠(𝑡) moving along the coordinate line. The objects initial displacement is zero 𝑣(𝑡) = 2𝑡 ℎ (𝑡 𝑘 − 2)4 Answer: 𝑠(𝑡) is nothing but the integration of 𝑣(𝑡) over time. That is, 𝑠(𝑡) = ∫ 𝑣(𝑡)𝑑𝑡 And intial displacement is given to be 0, which implies, 𝑠(0) = 0 On taking ℎ = 2 and 𝑘 = 3 we have 𝑠(𝑡) = ∫ 2𝑡 2 (𝑡 3 − 2)4 𝑑𝑡 Now we will be solving the above integral : 𝑠(𝑡) = ∫ 2𝑡 2 (𝑡 3 − 2)4 𝑑𝑡 ⇒ 𝑠(𝑡) = 2 ∫ 𝑡 2 (𝑡 3 − 2)4 𝑑𝑡 We will take below substitution: 𝑡 3 − 2 = 𝑢 ⇒ 3𝑡 2 𝑑𝑡 = 𝑑𝑢 ⇒ 𝑡 2 𝑑𝑡 = 𝑑𝑢 3 𝑡 3 − 2 = 𝑢 ⇒ 3𝑡 2 𝑑𝑡 = 𝑑𝑢 ⇒ 𝑡 2 𝑑𝑡 = 𝑑𝑢 3 The integral becomes: 𝑠(𝑡) = 2 ∫ 𝑢3 𝑑𝑢 3 ⇒ 𝑠(𝑡) = 2 ∫ 𝑡 2 (𝑡 3 − 2)4 𝑑𝑡 ⇒ 𝑠(𝑡) = 2𝑢5 +𝐶 15 Putting back the value of u, we get 𝑠(𝑡) = 2(𝑡 3 − 2)5 +𝐶 15 where C is a constant Now 𝑠(0) = 0 This implies, 𝑠(0) = 2(02 − 2)5 +𝐶 =0 15 ⇒0= 2.22 +𝐶 15 ⇒0= 26 +𝐶 15 ⇒− 26 =𝐶 15 ⇒− 64 =𝐶 15 Therefore, finally, we get 𝑠(𝑡) = 2(𝑡 3 − 2)5 15 Question #2 Find the general solution of the given differential equation 𝑥 𝑑𝑦 − 𝑦 = 𝑥 2 sin 𝑥 𝑑𝑥 give the largest interval over which the genreral solution is defined. Answer: 𝑥 𝑑𝑦 − 𝑦 = 𝑥 2 sin 𝑥 𝑑𝑥 ⇒ 𝑑𝑦 𝑦 − = 𝑥 sin 𝑥 𝑑𝑥 𝑥 This is 1-st order linear in y so integrating Factor = e 1 − dx x 1 = 𝑒 ln(𝑥) = 1 𝑥 multiplying the given equation with I.F. we get 1 𝑑𝑦 𝑦 − = sin 𝑥 𝑥 𝑑𝑥 𝑥 2 ⇒ 𝑑 𝑦 ( ) = sin 𝑥 𝑑𝑥 𝑥 integrating both side we get 𝑦 ∫ 𝑑 ( ) = sin 𝑥 𝑥 ⇒ 𝑦 = − cos 𝑥 + 𝐶 𝑥 ⇒ 𝑦 = −𝑥 cos 𝑥 + 𝐶𝑥 Since y is continuous on (−∞, ∞) so longest inter val of existence is (−∞, 𝑖𝑛𝑓𝑡𝑦) Since 𝑥 cos 𝑥 and 𝐶𝑥 both does not goes to 0 as 𝑥 → ∞ Hence then is no transient form. Question #3 What is the inverse function of 𝑓(𝑥) = 𝑥 2 ? Answer: Hence 𝑓(𝑥) = 𝑥 2 = 𝑥 2 ⇒ √𝑦 = √𝑥 2 ⇒ √𝑦 = |𝑥| = ± √𝑦 Question #4 How do you find the inverse of 𝑦 = (−𝑓𝑟𝑎𝑐12)𝑥 + 7? Answer: Explanation: To find the inverse of an equation, we switch x and y: 1 𝑥 = (− ) 𝑦 + 7 2 And solve for y: 1 𝑥 = (− ) 𝑦 + 7 (subtract 7 from both sides) 2 1 𝑥 − 7 = (− ) 𝑦 (multiply by 7 both sides) 2 2(𝑥 − 7) = 𝑦 𝑦 = 2𝑥 − 14 To write that this equation is the inverse of the original one, we write the expression as 𝑦 −1 = 2𝑥 − 14 Question #5 Sketch the graph of the following function 𝑓(𝑥) = −5𝑥 Use the graphing tool to graph the function Answer: Here is plot: FSP01511100702.jpgFSZ Question #6 Find the linear approximation of the function 𝑓(𝑥) = √4 − 𝑥 at 𝑎 = 0 and use it to approximate the numbers √3.9 and √3.99, please. Answer: Use the Taylor Form: 𝑓(𝑥) = 𝑓(𝑎) + (𝑥 − 𝑎) × 𝑓 ′ (𝑎) … ′ 𝑓(𝑥) = (√4 − 𝑥) = 1 −1 × 2 √4 − 𝑥 𝑓(0) = √4 − 0 = 2 𝑓 ′ (0) = 1 −1 1 × =− 2 √4 4 1 √3.9 = 𝑓(𝑥) ≈ 2 + 0.1 (− ) = 1.9750 4 1 √3.99 = 𝑓(𝑥) ≈ 2 + 0.01 (− ) = 1.9975 4 Question #7 A mass hanging from a vertical spring is in simple harmonic motion as given by the following position function, where t is measured in seconds and s is in inches: π 𝑠(𝑡) = 6 cos (π𝑡 + ). 4 Round answers to 2 decimal places. a. Determine the position of the spring at 𝑡 = 2.5 𝑠 b. Find the velocity of t spring 𝑡 = 2.5 𝑠 Answer: Step 1 The given position function is: π 𝑠(𝑡) = −6 cos (π𝑡 + ) 4 where t is in seconds and s(t) is in inches. (a) To determine 𝑠(2.5). That is position at 𝑡 = 2.5 So, π 𝑠(2.5) = −6 cos (π(2.5) + ) 4 = −6 cos ( 5π π + ) 2 4 = −6 cos ( 11π ) 4 ≈ 4.24 Question #8 Given the following statements: I. II. III. IV. The function The function The function The function 𝑓(𝑥) = 𝑥 2 sin2 𝑥 is an odd function. 𝑓(𝑥) = 𝑒 −𝑥 sin 𝑥 is a neither function. 𝑓(𝑥) = cos(5𝑥) sin2 𝑥 is an even function 𝑓(𝑥) = (𝑥 − 5)𝑒 2𝑥 is an even function. Answer: Step 1 we know that: put 𝑥 = −𝑥 and then, f (− x) = f ( x) → even function 𝑓(−𝑥) = −𝑓(𝑥) odd function 𝑓(−𝑥) ≠ −𝑓(𝑥) or f ( x) → neither function Step 2 The function 𝑓(𝑥)𝑥 2 sin2 𝑥 is odd function So, put 𝑥 = −𝑥 in function: 𝑓(−𝑥) = (−𝑥)2 (sin(−𝑥))2 = 𝑥 2 (−𝑠𝑖𝑛)2 (−𝑥) = 𝑥 2 (𝑠𝑖𝑛)2 , 𝑓(−𝑥) − 𝑓(𝑥) [this is a even function not odd function] (statement false) Question #9 1  m  x sin Let  x 0 x0 x=0 Find the set of values of m for which f(x) is continuous at 𝑥 = 0 f(x) is differentiable at 𝑥 = 0 f(x) is continuous but not differentiable at 𝑥 = 0. (i) (ii) (iii) Answer: The given function is: 1  m  x sin x  0 x0 x=0 f is continuous at 𝑥 = 0 if 𝑓(0) = lim 𝑓 (𝑥) 𝑥→0 Now, lim 𝑓 (𝑥) 𝑥→0 = lim x m sin x →0 1 x 𝑥 = 0 is 𝑚 > 0 lim 𝑓 (0) ≠ 0 𝑖𝑓 𝑚 < 0 𝑥→0 δ0 𝑓 is continuous for 𝑚𝑡(0, ∞) Question #10 Assume f and g are differentiable on their domains with ℎ(𝑥) = (𝑓𝑜𝑔)(𝑟). Suppose the equation of the line tangent to the graph of g at the point (4,7) is 𝑦 = 3𝑥 – 5 and the equation of the line tangent to the graph of f at (7,9) is 𝑦 = −2𝑥 + 23. (a) Calculate ℎ(4) and ℎ′ (4). (b) Determine an equation of the line tangent to the graph of h at the point on the graph where 𝑥 = 4. Answer: Step 1: Define the functions (a) Since ℎ(4) = 𝑓(𝑔(4)) and ℎ′ (4) = 𝑓 ′ (𝑔(4)) ⋅ 𝑔′ (4) , therefore, to determine the value of ℎ(4) and ℎ′ (4) : we need to know the values of 𝑔(4), 𝑓(𝑔(4)), 𝑔′ (4) and 𝑓 ′ (𝑔(4)) . Now, (4,7)point lies on the graph of g, therefore, 𝑔(4) = 7. Similarly, (7,9) lies on graph f , therefore, 𝑓(7) = 9. Now, 𝑔(4) = 7 , therefore, 𝑓(𝑔(4)) = 9. We know that if a line is tangent to a graph at a point (𝑥, 𝑦), then differentiation of the function will have the same value to the slope of the line. Now, slope of the line : 𝑦 = 3𝑥 − 5 is 3. As the the line 𝑦 = 3𝑥 − 5 is tangent to the graph g on (4,7), therefore, 𝑔′ (4) = 3. Similarly, we have, slope of the line 𝑦 = −2𝑥 + 23 is −2 . Therefore, 𝑓 ′ (7) = −2. Therefore, 𝑓 ′ (𝑔(4)) = −2. So, ℎ(4) = 𝑓(𝑔(4)) = 9. And, ℎ′ (4) = 𝑓 ′ (𝑔(4)). 𝑔′ (4) = −2.3 = −6. Question #11 Expand each function (using the appropiate technique/formula) Compute the derivative of the expanded function by applying the differentiation rules 𝑓(𝑥) = (𝑥 + 5)2 𝑓(𝑥) = (4𝑥 2 − 3)2 Answer: Step 1: To determine Derivative of the given functions by expanding them: 1) 𝑓(𝑥) = (𝑥 + 5)2 2) 𝑓(𝑥) = (4𝑥 2 − 3)2 Step 2: Formula used 1. (𝑥 + 𝑦)2 = 𝑥 2 + 2𝑥𝑦 + 𝑦 2 2. (𝑥 − 𝑦)2 = 𝑥 2 − 𝑥𝑦 + 𝑦 2 3. 𝑑 𝑑𝑥 (𝑥 𝑛 ) = 𝑛𝑥 𝑛−1 Step 3: Solution Consider the given function: 𝑓(𝑥) = (𝑥 + 5)2 Using formula, the expanded function is given by: 𝑓(𝑥) = 𝑥 2 + 2. 𝑥. 5 + 52 ⇒ 𝑓(𝑥) = 𝑥 2 + 10𝑥 + 25 Differentiating the above function with respect to x, we get, 𝑑𝑓(𝑥) 𝑑 2 (𝑥 + 10𝑥 + 25) = 𝑑𝑥 𝑑𝑥 ⇒ 𝑑𝑓(𝑥) 𝑑 2 𝑑 𝑑 (𝑥 ) + (10𝑥) + (25) = 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 ⇒ 𝑑𝑓(𝑥) = 2𝑥 + 10(1) + 0 𝑑𝑥 ⇒ 𝑑𝑓(𝑥) = 2𝑥 + 10 𝑑𝑥 Hence, the derivative of the given function is 𝑑𝑓 𝑑𝑥 = 2𝑥 + 10 Question #12 a) Consider the following functional relation, f, defined as: 𝑓: 𝑅 → 𝑅, 𝑓(𝑥) = 𝑥 2 Determine whether or not fis a bijection. If it is, prove it. If it is not, show why it is not b) Consider the set 𝐹 = 𝑦|𝑦 = 𝑎𝑥 3 + 𝑏 a, b being constants such that 𝑎 ≠ 0 and 𝑥 ∈ 𝑅. Is 𝐹 = 𝑅? If so, prove it. If not, explain in details why it is not the case. Answer: Step 1 (a). Given: 𝑓: 𝑅 → 𝑅, 𝑓(𝑥) = 𝑥 2 To check: f is one - one: 𝑓(𝑥) = 𝑓(𝑦) ⇒ 𝑥 = 𝑦 Here 𝑥 2 = 𝑦 2 ⇒ 𝑥 = ±𝑦 ⇒ 𝑓 is not one – one Step 2 To check: f is onto ∃𝑦 ∈ 𝑅 such that 𝑓(𝑥) ≠ 𝑦, ∀𝑥 ∈ 𝑅 Reason: suppose 𝑦 = −1 ∈ 𝑅 such that 𝑥 2 ≠ −1(𝑜𝑟)𝑥 2 = −1 ∈ 𝑅 Therefore f is not onto and hence f is not bijective Question #13 Consider the following. A= −5 9 −1 1 List the eigenvalues of A and bases of the corresponding eigenspaces. (Repeated eigenvalues should be entered repeatedly with the same eigenspaces.) smaller lambda-value λ1 = ___ smaller lambda-value λ2 = ___ Find an invertible matrix P and a diagonal matrix D such that 𝑝 −1 𝐴𝑃 = 𝐷. (Enter each matrix in the form [[𝑟𝑜𝑤1], [𝑟𝑜𝑤2], … ], where each row is a comma-separated list. If A is not diagonalizable, enter NO SOLUTION.) (𝐷, 𝑃) =(___) Answer: Step 1 Pms:- Given,Matrix A= −5 9 −1 1 To find Eigen Values & Eigen Vectors:EigenValues:-  −5 − d 9   | A − dz |=  −1 1− d   ⇒ (5 − 𝑑)(1 − 𝑑) + 9 = 0 ⇒ −5 + 5𝑑 − 𝑑 + 𝑑 2 + 9 = 0 ⇒ 𝑑 2 + 4𝑑 + 4 = 0 ⇒ (𝑑 + 2)2 = 0 ⇒ 𝑑 = −2, −2 Step 2 EigemVectors:-   −3 9   n1  0    [ A = 2 z ]x = 0     =    −1 3  n2  0      −3x1 + 9 x2 = 0 − x1 + 3x2 = 0 − x1 + 3 x2 = 0 x1 = 3 x2 Let 𝑥2 = 1 ⇒ 𝑥1 = 3 3  (V1 =   ) 1 So, 3 1  EigenValue 𝑑1 = −2; Eigenspace spam   3 1  EigenValue 𝑑2 = −2; Eigenspace spam   Step 3 Since, The number of EigenVectors is less than dimwnsions of the matrix, then the matrixis not diagonalizable. Hence, The matrix is not diagonolizable Ans:-(B) → No Since, The matrix is not diagonalizable. So, (𝐷, 𝑃) = (No solution)  −3 x + 9 x2 = 0   1   − x1 + 3x2 = 0   − x + 3x2 = 0   1   x1 = 3x2  Question #14 If ℎ(𝑥) = 5 for the function ℎ(𝑥) = 2𝑥 + 1, what is the value of x? Answer: We need to equate these two results of h(x) and solve for x: 2𝑥 + 1 = 5 2𝑥 + 1 − 1 = 5 − 1 2𝑥 + 0 = 4 2𝑥 = 4 2𝑥 2 = 4 2 2x =2 2 𝑥 = 2 Question #15 If 𝑓(𝑥) = 5𝑥 + 12, what is 𝑓 − 1(𝑥)? Answer: Let's assume f(x) is y for the sake of argument: 𝑦 = 5𝑥 + 12 Swap x and y and rerun the y-solve to find the inverse of the function. 𝑥 = 5𝑦 + 12 Subtract 12 on both sides: 𝑥 − 12 = 5𝑦 Divide by 5: 𝑦= 𝑥–12 5 Then 𝑓 −1 𝑥 = 𝑥−12 5 Question #16 How to find (f) if 𝑓 ′ (𝑥) = 16𝑥 3 + 6𝑥 + 7 and 𝑓(1) = −1 ? Answer: Integrating both sides comes first. 𝑓(𝑥) = ∫ 16𝑥 3 + 6𝑥 + 7𝑑𝑥 𝑓(𝑥) = 4𝑥 4 + 3𝑥 2 + 7𝑥 + 𝐶 We now determine C. −1 = 4(1)4 + 3(1)2 + 7(1) + 𝐶 −1 = 4 + 3 + 7 + 𝐶 −1 − 14 = 𝐶 𝐶 = −15 Question #17 Consider the functions 𝑓(𝑥) = 3𝑥 − 2 and 𝑔(𝑥) = 𝑥 2 − 1. What is the value of 𝑓[𝑔(5)]? Answer: We can first resolve g(5) We have 𝑔(𝑥) = 𝑥 2 − 1 Therefore 𝐺(5) = 52 − 1 𝐺(5) = 25 − 1 𝐺(5) = 24 Thus: 𝑓[𝑔(5)] = 𝑓[24] = (3 × 24) − 2 𝑓[𝑔(5)] = 72 − 2 𝑓[𝑔(5)] = 70 Question #18 For 𝑓(𝑥) = 24 − 2𝑥, how do you find f(2) and x such that 𝑓(𝑥) = 10? Answer: To calculate f(2), enter 𝑥 = 2 in place of f(x). 𝑓(2) = 24 − (2 × 2) 𝑓(2) = 24 − 4 𝑓(2) = 20 "given" 𝑓(𝑥) = 24 − 2𝑥" and " 𝑓(𝑥) = 10 "then we can equate the right sides" ⇒ 24 − 2𝑥 = 10 "subtract 24 from both sides" 24 −24 − 2x = 10 − 24 ⇒ −2𝑥 = −14 "divide both sides by - 2" −2 x −14 = −2 −2 ⇒= 7 Question #19 If 𝑔(𝑥) = 3(𝑥 − 2), what is the value of x if 𝑔(𝑥) = 6? Answer: 6 = 3(𝑥 − 2) Subtract 3 from both sides. 6 3 =𝑥−2 2=𝑥−2 2 more on each side 2+2=𝑥−2+2 4=𝑥+0 𝑥=4 Question #20 Given the function 𝑓(𝑥) = 𝑙𝑜𝑔3 (𝑥 + 4), what is the value of 𝑓 −1 (3)? Answer: We have, 𝑓(𝑥) = 𝑙𝑜𝑔3 (𝑥 + 4) We must first find the logarithm's inverse. Rewrite the logarithm in exponential form to accomplish this. 𝑦 = 𝑙𝑜𝑔3 (𝑥 + 4) 3𝑦 = 𝑥 + 4 Swap the positions of x and y. 3𝑥 = 𝑦 + 4 Solve for y. 𝑦 = 3𝑥 − 4 Substitute 𝑥 = 3. 𝑦 = 33 − 4 Solve for y. 𝑦 = 27 − 4 𝑦 = 23 𝑓 −1 (𝑥) = 23