Solution Manual For Principles of Taxation for Business and Investment Planning 2024 27th

Solution Manual For Principles of Taxation for Business and Investment Planning 2024 27th Edition By Sally Jones, Shelley RhoadesCatanach, Callaghan, Kubick (All Chapters, 100% Original Verified, A+ Grade) Find the right-sided Riemann Sum of sin(x^2) with n = 4, from [0, 4] - ANS-Right S4 = (1)[sin(1^2) + sin(2^2) + sin(3^2) + sin(4^2)] Right S4 = (1)[sin(1) + sin(4) + sin(9) + sin(16)] Right S4 = .201 Find the midpoint Riemann Sum of sin(x^2) with n = 4, from [0, 4] - ANS-Mid S4 = (1)[sin(.5^2) + sin(1.5^2) + sin(2.5^2) + sin(3.5^2)] Mid S4 = (1)[sin(.25) + sin(2.25) + sin(6.25) + sin(12.25)] Mid S4 = .681 Find the left-sided Riemann Sum of ln(x^2) with n=2, from [1, 3] - ANS-Left S2 = (1)[ln(1^2) + ln(2^2)] Left S2 = (1)[ln(1) + ln(4)] Left S2 = 1.386 Find the right-sided Riemann Sum of cos(x^2) with n = 3, from [2, 5] - ANS-Right S3 = (1)[cos(3^2) + cos(4^2) + cos(5^2)] Right S3 = (1)[cos(9) + cos(16) + cos(25)] Right S3 = -.878 Approximate the area between the x-axis and h(x) = 1/(7-x) from x = 2 to x = 5 using a left Riemann sum with 3 equal subdivisions. - ANS-Area = (1)[h(2) +h(3) +h(4)] Area = (1)[1/5 +1/4 +1/3] Area = 47/60 Use the trapezoidal approximation for the integral of (sinx)^2dx from [0, 1] with n = 4 to three decimal places. - ANS-Trapezoid = (1/2)(1/4)[(sin(0))^2 +2(sin(1/4))^2 +2(sin(1/2))^2 +2(sin(3/4))^2 +(sin(1))^2] Trapezoid = .277 Find the midpoint Riemann Sum of cos(x^2) with n = 4, from [0, 2] - ANS-Mid S4 = (1)(1/2)[cos(.25^2) + cos(.75^2) + cos(1.25^2) + cos(1.75^2)] Mid S4 = (1)(1/2)[cos(.625) + cos(.5625) + cos(1.5625) cos(3.0625)] Mid S4 = .824 If the function f is continuous for all real numbers and if f(x) = (x^2-7x +12)/(x -4) when x ≠ 4 then f(4) = - ANS-Factor numerator so f(x) = (x-3)(x-4)/(x-4) = x-3 f(4)=4-3 f(4) = 1 If f(x) = (x^2+5) if x 2, & f(x) = (7x -5) if x ≥ 2 for all real numbers x, which of the following must be true? I. f(x) is continuous everywhere. II. f(x) is differentiable everywhere. III. f(x) has a local minimum at x = 2. - ANS-At f(2) both the upper and lower piece of the discontinuity is 9 so the function is continuous everywhere. At f'(2) the upper piece is 4 and lower piece is 7 so f(x) is not differentiable everywhere. Since the slopes of the function on the left and right are both positive the function cannot have a local minimum or maximum at x= 2. Only I is true. For the function f(x) = (ax^3-6x), if x ≤ 1, & f(x) = (bx^2+4), x 1 to be continuous and differentiable, a = ..... - ANS-for the function to be continuous f(1) has to equal f(1): a(1^3) -6(1) = b(1^2) +4 a -6 = b +4 b=a-10 for the functions to be differentiable f'(1) has to equal f'(1): 3a(1^2) -6 = 2b(1) 3a -6 = 2b plug b from the first equation in to find a: 3a -6 = 2(a -10) a = -14 Find k if f(x) = (k) at x = 4 and f(x) = ((x^2 -16)/(x-4)) - ANS-1. f(4) exists and is equal to 8 2. lim from the left and right are both 8 3. lim f(x) as x approaches 4 is 8 which equals f(4) k must equal 8 If f(x) is continuous and differentiable and f(x) = (ax^4 +5x) for x ≤ 2, & f(x)= (bx^2 -3) for x 2 , then b =... - ANS-Plug x = 2 into both pieces. f(x) = (16a +10) for x ≤ 2, & (4b -6) for x 2 They must be equal to be continuous 16a +10 = 4b -6 a=.25b-1 Take the derivative of both pieces of this function and plug in x = 2 f(x) = (32a +5) for x ≤ 2, & f(x) = (4b -3) for x 2 They must be equal to be differentiable 32a +5 = 4b -3 plug in the first equation to find b 32(.25b-1)+5= 4b-3