Functions #15 - Questions and Answers | Carleton College

University:
Carleton College
Course:
MATH 120 | Calculus 2
Academic year:

2024
Views:

61

Pages:

2
Author:

Nevaeh Bauer

Question 1 Solve over the interval [0, 2π) 2(cos²x) + cosx = 0 Answer The terms of 2(cos²x) + cosx = 0 have a common factor of cos x. Factoring out cos x then gives cosx(2cosx + 1) = 0. The Zero Product Property states that if ab=0, then a=0 or b=0. Therefore, if cosx(2cosx + 1) = 0, then cos x = 0 or 2cosx + 1 = 0. From the unit circle, cos x = 0 when x = π/2 and 3π/2 2cosx + 1 = 0 when cosx = -1/2 , which is when x = 2π/3 and 4π/3 So the values are x = π/2 , 3π/2 , 2π/3 and 4π/3 Question 2 √-8(√-3 - √-5) Answer We are given: √-8(√-3 - √-5) Recall that sqrt-a=isqrta so we rewrite as: = i√8(i√3 - i√5) = 2i√2(i√3 - i√5) Apply distributive property: = 2i²√6 - 2i²√10) Recall that i² = -1 so: = 2(-1)√6 - 2(-1)√10 = -2√6 + 2√10 = 2√10 - 2√6 Question 3 Find the absolute maximum and absolute minimum values of f on the given interval and state where those values occur: f(x) = x³-3x² - 9x + 25, [-5, 10] Answer Step 1 Given function is f(x) = x³-3x² - 9x + 25 To find absolute minimum and asbsolute maximum over the interval(-5,10) Step 2 We have f(x) = x³ - 3x² - 9x + 25 To find absolute maximum and absolute maximum value firstly we nee dto find critical points differentiate the function respect to x Step 3 f'(x) = 3x² - 6x - 9 3x² - 6x - 9 by splitting middle term 3x² - 9x + 3x - 9 = 0 3x(x - 3) + 3(x - 3) = 0 (x - 3)(3x + 3) = 0 x - 3 = 0 or 3x + 3 = 0 x = 3 or 3x = -3 x = 3 or x = -1 Now plug x values in original equation f(3) = (3)³ - 3(3)² - 9(3) + 25 = 27 - 27 - 27 + 25 = -2 f(-1) = (-1)³ - 3(-1)² - 9(-1) + 25 = -1 - 3 + 9 + 25 = 30 Now check endpoints f(-5) = (-5)³ - 3(-5)² - 9(-5) + 25 = -125 - 75 + 45 + 25 = -130 f(10) = (10)³ - 3(10)² - 9(10) + 25 = 1000 - 300 - 90 + 25 = 635 Absolute max = 635 at x=10 and absolute min = -130 at x=-5 Question 4 What are the derivatives of the inverse trigonometric functions? How do the domains of the derivatives compare with the domains of the functions? Answer Step 1 To find the derivatives of inverse trignometric function and compare the domain of the function with the domain of the derivatives. Step 2 Derivatives of inverse Trignometric functions: (sin⁻¹x)' = 1/√(1-x²) (cos⁻¹x)' = -1/√(1-x²) (tan⁻¹x)' = 1/(1+x²) (cot⁻¹x)' = -1/(1+x²) (sec⁻¹x)' = 1/(|x|√(x²-1)) (cosec⁻¹x)' = -1/(|x|√(x²-1)) Step 3 Domain of the derivatives of inverse Trignometric functions: Function | Domain (sin⁻¹x)' | -1 < x < 1 (cos⁻¹x)' | -1 < x < 1 (tan⁻¹x)' | -∞ < x < ∞ (cot⁻¹x)' | -∞ < x < ∞ (sec⁻¹x)' | x ∈ (− ∞, −1) U (1, ∞) (cosec⁻¹x)' | x ∈ (− ∞, −1) U (1, ∞) Step 4 Domain of Inverse Trignometric Functions: Function | Domain sin⁻¹x | -1 ≤ x ≤ 1 cos⁻¹x | -1 ≤ x ≤ 1 tan⁻¹x | -∞ < x < ∞ cot⁻¹x | -∞ < x < ∞ sec⁻¹x | x ∈ (− ∞, −1] U [1, ∞) cosec⁻¹x | x ∈ (− ∞, −1] U [1, ∞) As we can conclude from the above tables that the domains of the derivatives of inverse trignometric functions are restrictive appropriately. 30.6s