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

2024
Views:

411

Pages:

4
Author:

Nevaeh Bauer

Question 1 Show that the inverse of any linear function f(x) = mx + b, where m ≠ 0, is also a linear function. Give the slope and y-intercept of the graph of f-1 in terms ofm and b. Answer y = mx + b x = my + b my = x - b y = (x/m) - (b/m) Slope: 1/m y-intercept: -b/m Question 2 Using a directrix of y = -2 and a focus of (2, 6), what quadratic function is created? f(x) = 1/16(x-2)² + 2 Answer To find equation of quadratic function Given, directrix, y= -2 And focus= (2,6) We know, from any point (x, y) on the parabola the focus and directrix are equidistant 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) and (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 y² + 36 - 12y - (y² + 4 + 4y) = -(x - 2)² y² + 36 - 12y - y² - 4 - 4y = -(x - 2)² -16y + 32 = -(x - 2)² (x - 2)² + 32 = 16y 16y = (x - 2)² + 32 y = 1/16 (x - 2)² + 32/16 y = 1/16 (x - 2)² + 2 Hence, the quadratic function is f(x) = 1/16(x - 2)² + 2 Question 3 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+k² (2) NST Shortest distance of the given line from origin |4(0) + 5- 0| / √(4² + (-1)²) Put 4x + 5 - y = 0 Put x = 0, y = 0 d = 5 / √17 (3) from equation (2)&(3), √(h² + k²) = 5/√17 h² + k² = 25/17 (4) Step 2 Put k = 4h + 5 into equation (4) h² + (4h + 5)² = 25/17 h² + 16h² + 25 + 40h = 25/17 17h² + 40h + 25 = 25/17 289h² + 680h + 425 = 25 289h² + 680h + 400 = 0 Use sridharacharya method to find root, A = 289, B = 680, C = 400 h = (-B ± √(B² - 4AC))/(2A) h = (-680 ± √((680)² - 4 * 289 * 400))/(2 * 289) h = (-680 ± 0)/(2 * 289) = -680/(578) = -40/34 = -20/17 so, k = 4h + 5 = 4(-20/17) + 5 = -80/17 + 85/17 = 5/17 Question 4 The inverse sine, inverse cosine, and inverse tangent functions have the following domains and ranges. (a) The function sin⁻¹ has domain ______ and range ______ (b) The function cos⁻¹ has domain ______ and range ______ (c) The function tan⁻¹ has domain ______ and range ______ Answer (a)The inverse sine has a domain of [-1,1], and a range of [-π/2, π/2] (b) The inverse cosine has a domain of [-1,1], and a range of [0, π] (c) The inverse tangent has a domain of R, and a range if (-π/2, π/2) Result: (a) [-1,1], [-π/2, π/2] (b) [-1, 1], [0, π] (c) R, (-π/2, π/2) Question 5 A rectangle has area 16m². Express the perimeter of the rectangle as a function of the length of one of its sides. Answer Two variables are required to represent a rectangle's sides. Let's decide on x for width and y for length. Knowing that the rectangle's area is 16, we can now represent it. xy = 16 y = 16/x Perimeter of a rectangle. P = 2x + 2y P = 2x + 2(16/x) P = 2x + 32/x P = (2x² + 32)/x Note that the domain will be x>0 as the width must be positive. If the length is larger than the width then the domain will be 0 x => 16 > x² => x < 4. Answer: P = (2x² + 32) / x