Practice Test - Chapter 1

Transcription

1 Determine whether the given relation represents y as a function of x. 1. y 3 x = 5 When x = 1, y = ±. Therefore, the relation is not one-to-one and not a function. not a function 4. PARKING The cost of parking a car downtown is $0.75 per 30 minutes for a maximum of $4.50. Parking is charged per second. a. Write a function for c(x), the cost of parking a car for x hours. b. Find c(2.5). c. What is the domain for c(x)? Explain your reasoning. a. A cost of $0.75 per 30 minutes is equal to a cost of $1.50 per hour. The charge maximizes at $4.50 so will remain unchanged when the car is parked for longer than 3 hours. c(x) = 2. The graph passes the Vertical Line Test, so the relation is a function. function 3. y = For every x-value, there is only one corresponding y- value. Therefore, the relation is one-to-one and a function. Also, the graph of the relation passes the Vertical Line Test. function 5. b = 3.75 c. D = [0, 3]; Sample answer: The number of hours must be greater than or equal to 0. a. c(x) = b. $3.75 c. D = [0, 3]; Sample answer: The number of hours must be greater than or equal to 0. State the domain and range of each function. The graph continues for all values of x and has a minimum y-value of 3. D = (, ), R = [ 3, ) D = (, ), R = [ 3, ) esolutions Manual - Powered by Cognero Page 1

2 6. The graph continues for all values of x such that x 5 and has a minimum y-value of 0. D = (, 5], R = [0, ) D = (, 5], R = [0, ) Find the y-intercept(s) and zeros for each function. 7. f (x) = 4x 2 8x MULTIPLE CHOICE Which relation is symmetric about the x-axis? A x 2 yx = 2 B x 3 y = 8 C y = x D y 2 = 4x In order for a relation to be symmetric with respect to the x-axis, (x, y) must correspond with (x, y). This occurs for choice D only. (x, y) (x, y) (0.25, 1) (0.25, 1) (1, 2) (1, 2) (2.25, 3) ( 2.25, 3) (4, 4) ( 4, 4) D Determine whether each function is continuous at x = 3. If discontinuous, identify the type of discontinuity as infinite, jump, or removable ; 1, 3 8. f (x) = x 3 + 4x 2 + 3x f(3) = 6,, The function is continuous at x = 3. continuous 11. f (x) = 0; 3, 1, 0 not continuous; removable discontinuity at x = 3 not continuous; removable discontinuity esolutions Manual - Powered by Cognero Page 2

3 Find the average rate of change for each function on the interval [ 2, 6]. 12. f (x) = x 4 + 3x Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing or decreasing f (x) = 14. f is increasing on (, 2.5) and decreasing on (2.5, ). f is increasing on (, 2.5) and decreasing on (2.5, ). 15. f is decreasing on (, 1.5), increasing on ( 1.5, 0), decreasing on (0, 1.5), and increasing on (1.5, ). f is decreasing on (, 1.5), increasing on ( 1.5, 0), decreasing on (0, 1.5), and increasing on (1.5, ). esolutions Manual - Powered by Cognero Page 3

4 16. Which function is shown in the graph? Identify the parent function f (x) of g(x). Then sketch the graph of g(x). 17. g(x) = (x + 3) 3 f(x) = x 3 F f (x) = x 4 3 G f (x) = x H f (x) = x J f (x) = x The parent function g(x) = x is shifted down 3 units. Therefore, F and H are possible choices. The graph is also shifted 4 units left, which corresponds to choice H. f(x) = x 3 H esolutions Manual - Powered by Cognero Page 4

5 18. g(x) = x 2 4 f(x) = x Given f (x) = x 6 and g(x) = x 2 36, find each function and its domain. f(x) = x 2 The value of x cannot equal 6 or 6. In the original expression, if x = 6 or 6, a division by 0 would occur, which is undefined. for x 6 or x [g f ](x) There are no restrictions on the domain. [f g](x) = x 2 12x esolutions Manual - Powered by Cognero Page 5

6 21. TEMPERATURE In most countries, temperature is measured in degrees Celsius. The equation that relates degrees Fahrenheit with degrees Celsius is F = C a. Write C as a function of F. b. Find two functions f and g such that C = [f g] (F). a. Determine whether f has an inverse function. If it does, find the inverse function and state any restrictions on its domain. 22. f (x) = (x 2) 3 Graph f (x). b. When making a composition, look for the operation that must be done first. This will represent the innermost function in the composition, in this The graph passes the Horizontal Line Test. Therefore, an inverse exists. case, g(x). Sample answer: (x) = x; g(x) = x 32 a. C = (F 32) b. Sample answer: f (x) = x; g(x) = x 32 There is no domain restriction. yes; f 1 (x) = + 2 esolutions Manual - Powered by Cognero Page 6

7 23. f (x) = Graph f (x). 24. f (x) = Graph f (x). The graph passes the Horizontal Line Test. Therefore, an inverse exists. The graph passes the Horizontal Line Test. Therefore, an inverse exists. Graph the inverse. Graph the inverse. The range of f (x) = [0, ), so the domain of f 1 (x) must be limited to [0, ). The domain of f 1 (x) is x 1. yes; f 1 = 4 x 2 ; x 0 yes; f 1 (x) = ; x 1 esolutions Manual - Powered by Cognero Page 7

8 25. f (x) = x 2 16 Graph f (x). The graph does not pass the Horizontal Line Test. Therefore, an inverse does not exist. no esolutions Manual - Powered by Cognero Page 8