#$&* course Mth 158 12/3/12 7:13 pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
.............................................
Given Solution: * * A linear function, represented most simply by y = x, has no curvature. A quadratic function, represented most simply by y = x^2, has a parabolic graph, which is either concave up or concave down. A cubic function, represented most simply by y = x^3, has a graph which changes concavity at the origin. This is the first power which can change concavity. A square root function, represented most simply by y = sqrt(x), have a graph which consists of half of a parabola opening to the right or left. Its concavity does not change. A reciprocal function, represented most simply by y = 1 / x, has a graph with a vertical asymptote at the origin and horizontal asymptotes at the x axis, and opposite concavity on either side of the asymptote. An absolute value function, represented most simply by y = | x |, has a graph which forms a 'V' shape. The greatest integer function [[ x ]] is a 'step' function whose graph is made up of horizontal 'steps'. The concavity of the reciprocal function changes, as does the concavity of the cubic function. The cubic function is the only one that matches the graph, which lacks the vertical and horizontal asymptotes. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 3.4.20 (was 3.3.12). Does your sketch of f(x) = sqrt(x) increase or decrease, and does it do so at an increasing or decreasing rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I substituted the values sqrt(0), sqrt(2) and sqrt(4) at x = 0, 2 and 4. I came up with the following: sqrt(0) = 0, sqrt(2) = 1.414 approx., and sqrt(4) = 2 The values increase therefore the graph increases with each value of x. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: * * y = sqrt(x) takes y values sqrt(0), sqrt(2) and sqrt(4) at x = 0, 2 and 4. sqrt(0) = 0, sqrt(2) = 1.414 approx., and sqrt(4) = 2. The change in y from the x = 0 to the x = 2 point is about 1.414; the change from the x = 2 to the x = 4 point is about .586. The y values therefore increase, but by less with each 'jump' in x. So the graph contains points (0,0), (2, 1.414) and (4, 2) and is increasing at a decreasing rate. ** What three points did you label on your graph? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I feel like I understood the problem but could have elaborated more. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: * 3.4.34 / 24 (was 3.3.24) f(x) = 2x+5 on (-3,0), -3 at 0, -5x for x>0. Given the intercepts, domain and range of the function YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I am a bit confused on this problem, but this is what I get. The graph of y = 2x + 5 has y intercept of 5 and slope of 2, and the x intecept occurs when y = 0. The x intercept occurs at 2x + 5 = 0 or x = -5/2. The graph should include the point (0, -3).
.............................................
Given Solution: * * From x = -3 to x = 0 the graph coincides with that of y = 2 x + 5. The graph of y = 2x + 5 has y intercept 5 and slope 2, and its x intecept occurs when y = 0. The x intercept therefore occurs at 2x + 5 = 0 or x = -5/2. Since this x value is in the interval from -3 to 0 it is part of the graph. At x = -3 we have y = 2 * (-3) + 5 = -1. So the straight line which depicted y = 2x + 5 passes through the points (-3, -1), (-5/2, 0) and (0, 5). The part of this graph for which -3 < x < 0 starts at (-3, -1) and goes right up to (0, 5), but does not include (0, 5). For x > 0 the value of the function is -5x. The graph of -5x passes through the origin and has slope -5. The part of the graph for which x > 0 includes all points of the y = -5x graph which lie to the right of the origin, but does not include the origin. For x = 0 the value of the function is given as -3. So the graph will include the single point (0, -3). ** STUDENT COMMENT I cant get a grasp on the domain and range. I've studied it in the book its just not sinking in yet. INSTRUCTOR RESPONSE The function is defined for x values on the interval (-3, 0), at x = 0 and for x > 0. So the function is defined for all x values which are greater than 3. Thus the domain of this function in the interval (-3, infinity). The range is the set of all possible values of f(x), where x can be any number in the domain. On the interval from x = -3 to x = 0 the function coincides with the function y = 2 x + 5. A graph of y = 2 x + 5 would be linear, increasing at a constant rate from 2 * (-3) + 5 = -1 at x = -3 to 2 * 0 + 5 = 5 at x = 0. Our function f(x) is not equal to 2 x + 5 at x = -3 (where it is not defined), and at x = 0 it is assigned the value -3. However for all x values between -3 and 0 our function is equal to y = 2 x + 5, so it takes every value between -1 and 5. For x > 0 the function coincides with y = - 5 x. This is a linear function, decreasing at a constant rate, whose value at x = 0 is 0. For x > 0 this linear function takes every possible negative value. Our function f(x) therefore takes all values between -1 and 5, as well as all possible negative values. Thus is takes all values from -infinity to 5, not including 5. So its range is (-infinity, 5). Note that f(x) also takes the value -3 at x = 0, but that is already in the range (-infinity, 5) and doesn't add anything to the range. The graph below depicts the linear functions y = 2x + 5, y = -5x and the point (0, -3): The next figure depicts the function f(x) as the heavy green line segments and the dot over the point (0, -3): The domain and range of the function are indicated by the red and blue rays on the x and y axes, respectively. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): After reading your response it does make more sense. In order to get this down, I will definitely need more practice. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: * 3.4.34 / 24 (was 3.3.24) f(x) = 2x+5 on (-3,0), -3 at 0, -5x for x>0. Given the intercepts, domain and range of the function YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I am a bit confused on this problem, but this is what I get. The graph of y = 2x + 5 has y intercept of 5 and slope of 2, and the x intecept occurs when y = 0. The x intercept occurs at 2x + 5 = 0 or x = -5/2. The graph should include the point (0, -3).
.............................................
Given Solution: * * From x = -3 to x = 0 the graph coincides with that of y = 2 x + 5. The graph of y = 2x + 5 has y intercept 5 and slope 2, and its x intecept occurs when y = 0. The x intercept therefore occurs at 2x + 5 = 0 or x = -5/2. Since this x value is in the interval from -3 to 0 it is part of the graph. At x = -3 we have y = 2 * (-3) + 5 = -1. So the straight line which depicted y = 2x + 5 passes through the points (-3, -1), (-5/2, 0) and (0, 5). The part of this graph for which -3 < x < 0 starts at (-3, -1) and goes right up to (0, 5), but does not include (0, 5). For x > 0 the value of the function is -5x. The graph of -5x passes through the origin and has slope -5. The part of the graph for which x > 0 includes all points of the y = -5x graph which lie to the right of the origin, but does not include the origin. For x = 0 the value of the function is given as -3. So the graph will include the single point (0, -3). ** STUDENT COMMENT I cant get a grasp on the domain and range. I've studied it in the book its just not sinking in yet. INSTRUCTOR RESPONSE The function is defined for x values on the interval (-3, 0), at x = 0 and for x > 0. So the function is defined for all x values which are greater than 3. Thus the domain of this function in the interval (-3, infinity). The range is the set of all possible values of f(x), where x can be any number in the domain. On the interval from x = -3 to x = 0 the function coincides with the function y = 2 x + 5. A graph of y = 2 x + 5 would be linear, increasing at a constant rate from 2 * (-3) + 5 = -1 at x = -3 to 2 * 0 + 5 = 5 at x = 0. Our function f(x) is not equal to 2 x + 5 at x = -3 (where it is not defined), and at x = 0 it is assigned the value -3. However for all x values between -3 and 0 our function is equal to y = 2 x + 5, so it takes every value between -1 and 5. For x > 0 the function coincides with y = - 5 x. This is a linear function, decreasing at a constant rate, whose value at x = 0 is 0. For x > 0 this linear function takes every possible negative value. Our function f(x) therefore takes all values between -1 and 5, as well as all possible negative values. Thus is takes all values from -infinity to 5, not including 5. So its range is (-infinity, 5). Note that f(x) also takes the value -3 at x = 0, but that is already in the range (-infinity, 5) and doesn't add anything to the range. The graph below depicts the linear functions y = 2x + 5, y = -5x and the point (0, -3): The next figure depicts the function f(x) as the heavy green line segments and the dot over the point (0, -3): The domain and range of the function are indicated by the red and blue rays on the x and y axes, respectively. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): After reading your response it does make more sense. In order to get this down, I will definitely need more practice. ------------------------------------------------ Self-critique Rating: 2 #*&! ********************************************* Question: * 3.4.34 / 24 (was 3.3.24) f(x) = 2x+5 on (-3,0), -3 at 0, -5x for x>0. Given the intercepts, domain and range of the function YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I am a bit confused on this problem, but this is what I get. The graph of y = 2x + 5 has y intercept of 5 and slope of 2, and the x intecept occurs when y = 0. The x intercept occurs at 2x + 5 = 0 or x = -5/2. The graph should include the point (0, -3).
.............................................
Given Solution: * * From x = -3 to x = 0 the graph coincides with that of y = 2 x + 5. The graph of y = 2x + 5 has y intercept 5 and slope 2, and its x intecept occurs when y = 0. The x intercept therefore occurs at 2x + 5 = 0 or x = -5/2. Since this x value is in the interval from -3 to 0 it is part of the graph. At x = -3 we have y = 2 * (-3) + 5 = -1. So the straight line which depicted y = 2x + 5 passes through the points (-3, -1), (-5/2, 0) and (0, 5). The part of this graph for which -3 < x < 0 starts at (-3, -1) and goes right up to (0, 5), but does not include (0, 5). For x > 0 the value of the function is -5x. The graph of -5x passes through the origin and has slope -5. The part of the graph for which x > 0 includes all points of the y = -5x graph which lie to the right of the origin, but does not include the origin. For x = 0 the value of the function is given as -3. So the graph will include the single point (0, -3). ** STUDENT COMMENT I cant get a grasp on the domain and range. I've studied it in the book its just not sinking in yet. INSTRUCTOR RESPONSE The function is defined for x values on the interval (-3, 0), at x = 0 and for x > 0. So the function is defined for all x values which are greater than 3. Thus the domain of this function in the interval (-3, infinity). The range is the set of all possible values of f(x), where x can be any number in the domain. On the interval from x = -3 to x = 0 the function coincides with the function y = 2 x + 5. A graph of y = 2 x + 5 would be linear, increasing at a constant rate from 2 * (-3) + 5 = -1 at x = -3 to 2 * 0 + 5 = 5 at x = 0. Our function f(x) is not equal to 2 x + 5 at x = -3 (where it is not defined), and at x = 0 it is assigned the value -3. However for all x values between -3 and 0 our function is equal to y = 2 x + 5, so it takes every value between -1 and 5. For x > 0 the function coincides with y = - 5 x. This is a linear function, decreasing at a constant rate, whose value at x = 0 is 0. For x > 0 this linear function takes every possible negative value. Our function f(x) therefore takes all values between -1 and 5, as well as all possible negative values. Thus is takes all values from -infinity to 5, not including 5. So its range is (-infinity, 5). Note that f(x) also takes the value -3 at x = 0, but that is already in the range (-infinity, 5) and doesn't add anything to the range. The graph below depicts the linear functions y = 2x + 5, y = -5x and the point (0, -3): The next figure depicts the function f(x) as the heavy green line segments and the dot over the point (0, -3): The domain and range of the function are indicated by the red and blue rays on the x and y axes, respectively. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): After reading your response it does make more sense. In order to get this down, I will definitely need more practice. ------------------------------------------------ Self-critique Rating: 2 #*&!#*&! ********************************************* Question: * 3.4.34 / 24 (was 3.3.24) f(x) = 2x+5 on (-3,0), -3 at 0, -5x for x>0. Given the intercepts, domain and range of the function YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I am a bit confused on this problem, but this is what I get. The graph of y = 2x + 5 has y intercept of 5 and slope of 2, and the x intecept occurs when y = 0. The x intercept occurs at 2x + 5 = 0 or x = -5/2. The graph should include the point (0, -3).
.............................................
Given Solution: * * From x = -3 to x = 0 the graph coincides with that of y = 2 x + 5. The graph of y = 2x + 5 has y intercept 5 and slope 2, and its x intecept occurs when y = 0. The x intercept therefore occurs at 2x + 5 = 0 or x = -5/2. Since this x value is in the interval from -3 to 0 it is part of the graph. At x = -3 we have y = 2 * (-3) + 5 = -1. So the straight line which depicted y = 2x + 5 passes through the points (-3, -1), (-5/2, 0) and (0, 5). The part of this graph for which -3 < x < 0 starts at (-3, -1) and goes right up to (0, 5), but does not include (0, 5). For x > 0 the value of the function is -5x. The graph of -5x passes through the origin and has slope -5. The part of the graph for which x > 0 includes all points of the y = -5x graph which lie to the right of the origin, but does not include the origin. For x = 0 the value of the function is given as -3. So the graph will include the single point (0, -3). ** STUDENT COMMENT I cant get a grasp on the domain and range. I've studied it in the book its just not sinking in yet. INSTRUCTOR RESPONSE The function is defined for x values on the interval (-3, 0), at x = 0 and for x > 0. So the function is defined for all x values which are greater than 3. Thus the domain of this function in the interval (-3, infinity). The range is the set of all possible values of f(x), where x can be any number in the domain. On the interval from x = -3 to x = 0 the function coincides with the function y = 2 x + 5. A graph of y = 2 x + 5 would be linear, increasing at a constant rate from 2 * (-3) + 5 = -1 at x = -3 to 2 * 0 + 5 = 5 at x = 0. Our function f(x) is not equal to 2 x + 5 at x = -3 (where it is not defined), and at x = 0 it is assigned the value -3. However for all x values between -3 and 0 our function is equal to y = 2 x + 5, so it takes every value between -1 and 5. For x > 0 the function coincides with y = - 5 x. This is a linear function, decreasing at a constant rate, whose value at x = 0 is 0. For x > 0 this linear function takes every possible negative value. Our function f(x) therefore takes all values between -1 and 5, as well as all possible negative values. Thus is takes all values from -infinity to 5, not including 5. So its range is (-infinity, 5). Note that f(x) also takes the value -3 at x = 0, but that is already in the range (-infinity, 5) and doesn't add anything to the range. The graph below depicts the linear functions y = 2x + 5, y = -5x and the point (0, -3): The next figure depicts the function f(x) as the heavy green line segments and the dot over the point (0, -3): The domain and range of the function are indicated by the red and blue rays on the x and y axes, respectively. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): After reading your response it does make more sense. In order to get this down, I will definitely need more practice. ------------------------------------------------ Self-critique Rating: 2 #*&!#*&!#*&!