Multivariable functions #2 - Questions and Answers

Question 1 Select the correct answer: Which statement best describes the zeros of the function h(x) = (x-6) (x² + 8x + 16)? The function has two distinct real zeros The function has three distinct real zeros. The function has one real and two complex zeros. The function has three complex zeros. Answer Step 1 Given: f(x) = (x-6) (x² + 8x + 16) Step 2 Finding its zeroes, which equates the function to zero, gives us, (x-6) (x² + 8x + 16) = 0 The term (x² + 8x + 16) is a quadratic polynomial which can be factorized by split the middle term method proceeding accordingly, ⇒ (x-6) (x² + 4x + 4x + 16) = 0 ⇒ (x-6) [x(x+4) + 4(x+4)] = 0 ⇒ (x-6) [(x+4) (x+4)] = 0 Now, if (x-6) = 0 ⇒ x = 6 if (x + 4) = 0 ⇒ x = -4 ∴ zeroes of the function are 6 and -4. ∴ the function has two distinct real roots, i.e. option (1) is correct. Question 2 Find all the complex roots. Leave your answers in polar form with the argument in degrees. The complex fourth roots of 100 + 100√3i Answer Step 1 Consider the given complex number 100 + 100√3i step 2 Convert the complex number into polar form. We have that a=100 and b = 100√3 r = √(a² + b²) = √(100)² + (100√3)² = 200 θ = tan⁻¹(100√3 / 100) = π / 3 = 60° ∴ 100 + 100√3i = 200(cos 60° + i sin 60°) Step 3 Find the fourth roots of complex number. The nth roots of a complex number is given by zₖ = ⁿ√r [cos ((θ + 360°k) / n) + i sin ((θ + 360°k) / n)], k = 0, 1, …, n-1 We have n=4, r=200, θ = 60° zₖ = ⁴√200 [cos ((60° + 360°k) / 4) + i sin ((60° + 360°k) / 4)], k = 0,1,2,3 Therefore, zₖ = 2⁴√5 [cos (15° + 90°k) + i sin (15° + 90°k)], k = 0, 1, 2, 3 Question 3 Find all complex zeros of the polynomial function. Give exact values. List multiple zeros as necessary. f(x) = x⁴ - 5x³ - 36x² + 272x - 448 All complex zeros are=? Answer Step 1 Given, f(x) = x⁴ - 5x³ - 36x² + 272x - 448 Factorizing f(x) as follows: x⁴ - 5x³ - 36x² + 272x - 448 = x³(x - 4) - x²(x - 4) - 40x(x - 4) + 112(x - 4) = (x - 4) (x³ - x² - 40x + 112) = (x - 4)(x - 4) (x² + 3x - 28) = (x - 4)(x-4)(x - 4)(x + 7) = (x - 4)³(x + 7) Step 2 Equating it to zero (x - 4)³(x + 7) = 0 x - 4 = 0 x + 7 = 0 x = 4, x = -7 Thus, there are no complex zeros. Question 4 There are 100 two-bedroom apartments in the apartment building Lynbrook West.. The monthly profit (in dollars) realized from renting out x apartments is given by the following function. P(x) = -12x² + 2136x - 41000 How many units should be rented out in order to optimize the monthly rental profit? What is the maximum monthly profit realizable? Answer Step 1 Given profit function, P(x) = -12x² + 2136x - 41000 To maximize profit dP(x) / dx = 0 d/dx (-12x² + 2136x - 41000) = 0 -24x + 2136 = 0 -24x = -2136 x = -2136 / -24 x = 89 Here, d²P(x)/dx² = -24 < 0 as per second derivative test profit is maximum at x = 89 Also, maximum profit = P(89) = -12(89)² + 2136(89) - 41000 = 54052 To maximize the monthly rental profit, 89 units should be rented out, 54052$ be the maximum monthly profit Question 5 Find the complex zeros of the following polynomial function. Write f in factored form. f(x) = x³ - 12x² + 49x - 58 The complex zeros of f are =? Answer Step 1 Given: f(x) = x³ - 12x² + 49x - 58 It can be written as, f(x) = x³ - 2x² - 10x² + 20x + 29x - 58 f(x) = x²(x - 2) - 10x(x - 2) + 29(x - 2) f(x) = (x - 2)(x² - 10x + 29) Step 2 Roots of f(x) or zeroes of f(x) are given by (x-2)(x² - 10x + 29) ⇒ x = 2 or x² - 10x + 29 = 0 x = 2 or x = [10 ± √(100 - 4*29)] / 2 x = 2 or x = (10 ± √-16) / 2 x = 2 or x = 5 ± 2i (complex zeroes)