course Mth 158
......!!!!!!!!...................................
22:25:57 query 3.1.20 (was 3.1.6) (-2,5),(-1,3),(3,7),(4,12)}Is the given relation a function? Why or why not? If so what are its domain and range?
......!!!!!!!!...................................
RESPONSE -->
.................................................
Vr䖜vݰүS assignment #022 {g逷ẕ College Algebra 03-05-2006
......!!!!!!!!...................................
22:27:39 query 3.1.20 (was 3.1.6) (-2,5),(-1,3),(3,7),(4,12)}Is the given relation a function? Why or why not? If so what are its domain and range?
......!!!!!!!!...................................
RESPONSE --> d=(-2,-2,3,4) r=(5,3,7,12) one time
.................................................
......!!!!!!!!...................................
22:27:46 This relation is a function because every first element is paired with just one second element--there are no distinct ordered pairs with the same first element.
the domain is ( -2,-2,3,4) the range is ( 5,3,7,12) Another way of saying that this is a function is that every element of the domain appears only once in the relation.......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
22:29:23 query 3.1.34 (was 3.1.20) f(0), f(1), f(-1), f(-x), -f(x), f(x+1), f(2x), f(x+h) for 1 - 1 / (x+1)^2What are you expressions for f(0), f(1), f(-1), f(-x), -f(x), f(x+1), f(2x), f(x+h)?query 3.1.30. y = (3x-1)/(x+2)
......!!!!!!!!...................................
RESPONSE --> f(-x)=-(1,-1/9 (x+2)sqd) =-1+1/(x+2)sqd this is a function
.................................................
......!!!!!!!!...................................
22:29:42 STUDENT SOLUTION WITH INSTRUCTOR COMMENTS: f(x) = 1- 1/(x+2)^2
f(0) = 1- 1/ (0+2)^2 f(0) = 1-1/4 f(0) = 3/4 f(1) = 1- 1/ (3)^2 f(1) = 1- 1/9 f(1) = 8/9 f(-1) = 1- 1/(-1+2)^2 f(-1)= 1-1 f(-1)= 0 f(-x)= 1- 1/(-x+2)^2 f(-x)= 1 -1/ (x^2-4x+4) -f(x) = -(1- 1/(x+2)^2) -f(x)= -(1 - 1/ (x^2+4x+4)) -f(x) = (1/(x^2 + 4x + 4)) - 1 ** Your answer is right but you can leave it in factored form: f(-x) = -(1 - 1/(x+2)^2) = -1 + 1 / (x+2)^2. ** f(x+1) = 1- 1/((x+1) + 2)^2. This can be expanded as follows, but the expansion is not necessary at this point: = 1- 1/ ((x+1)^2 +2(x+1) + 2(x+1)+4) = 1- 1/ ((x+1)^2 +8x+8) = 1- 1/ (x^2+2x+1+8x+8) = 1- 1/(x^2 + 10x +9) ** Good algebra, and correct, but again no need to expand the square, though it is perfectly OK to do so. ** f(2x)= 1-1/(2x+2)^2 = 1- 1/(4x^2+8x+4) ** same comment ** f(x+h)= 1- 1/((x+h)+2)^2 = 1- 1/((x+h)^2 + 4(x+h) + 4) = 1- 1/ (x^2 + 2xh + h^2+4x+4h+4) ** same comment **......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
22:29:59 query 3.1.44 (was 3.1.30)
Is y = (3x-1)/(x+2) the equation of a function?......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
22:30:12 ** This is a function. Any value of x will give you one single value of y, and all real numbers x except -2 are in the domain. So for all x in the domain this is a function. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:31:25 query 3.1.54 (was 3.1.40). G(x) = (x+4)/(x^3-4x)
......!!!!!!!!...................................
RESPONSE --> denominator is x=0,2 or-2 all real but (0,2,-2)
.................................................
......!!!!!!!!...................................
22:31:33 ** Starting with
g(x) = (x+4) / (x^3-4x) we factor x out of the denominator to get g(x)= (x+4) / (x (x^2-4)) then we factor x^2 - 4 to get g(x) = (x+4) / (x(x-2)(x+2)). The denominator is zero when x = 0, 2 or -2. The domain is therefore all real numbers such that x does not equal {0,2,-2}. **......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
22:31:51 **** query 3.2.12 (was 3.1.50). Pos incr exp fnDoes the given graph depict a function? Explain how you determined whether or not the graph depicts a function.If the graph does depict a function then what are the domain and range of this function?
......!!!!!!!!...................................
RESPONSE --> verticle line test function
.................................................
......!!!!!!!!...................................
22:31:58 using the vertical line test we determine this is a function, since any given vertical line intersects the graph in exactly one point.
The function extends all the way to the right and to the left, and there are no breaks, so the domain consists of all real numbers. The range consists of all possible y values. The function takes all y values greater than zero so the range is {y | y>0}, expressed in interval notation as (0, infinity). The y intercept is (0,1); there is no x intercept but the negative x axis is an asymptote. This graph has no symmetery.......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
22:32:59 query 3.2.16 (was 3.1.54) Circle rad 2 about origin.
......!!!!!!!!...................................
RESPONSE --> no breaks all to left and right is the line r=-(y/y>0) d=all reals
.................................................
......!!!!!!!!...................................
22:33:12 Using the vertical line test we see that every vertical line lying between x = -2 and x = 2, not inclusive of x = -2 and x =2, intersects the graph in two points. So this is not a function
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
22:33:32 query 3.2.22 (was 3.1.60). Downward hyperbola vertex (1,2 5).Does the given graph depict a function? Explain how you determined whether or not the graph depicts a function.If the graph does depict a function then what are the domain and range of this function?
......!!!!!!!!...................................
RESPONSE --> not a function
.................................................
......!!!!!!!!...................................
22:33:41 Every vertical line intersects the graph at exacty one point so the graph depicts a function.
The function extends to the right and to the left without breaks so the domain consists of all real numbers. The range consists of all possible y values.......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
22:34:04 query 3.1.82 (was 3.1.70). f(x) =(2x - B) / (3x + 4).
If f(0) = 2 then what is the value of B?......!!!!!!!!...................................
RESPONSE --> d=all reals r all ys
.................................................
......!!!!!!!!...................................
22:34:11 If f(0) = 2 then we have
2 = (2 * 0 - B) / (3 * 0 + 4) = -B / 4, so that B = -4 * 2 = -8. if f(2)=1/2 what is value of B? If f(2) = 1/2 then we have 1/2 = ((2*2)-B) / ((3*2)+4) 1/2 = (4-B) / 10 5 = 4-B 1=-B B=-1 **......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
22:34:34 query 3.1.90 (was 3.1.80). H(x) = 20 - 13 x^2 (falling rock on Jupiter)What are the heights of the rock at 1, 1.1, 1.2 seconds?When is the rock at each altitude: 15 m, 10 m, 5 m.When does the rock strike the ground?
......!!!!!!!!...................................
RESPONSE --> too hard for me
.................................................
......!!!!!!!!...................................
22:34:42 GOOD STUDENT SOLUTION: The height at t = 1 is
H(1) = 20-13 H(1) = 7m The height at t = 1.1 is H(1.1)= 20-13(1.1)^2 = 20-13(1.21) = 20-15.73 H(1.1)= 4.27m. The height at t = 1.2 is H(1.2)= 20 - 13*(1.2)^2 = 20- 13 *(1.44) = 20-18.72 H(1.2) = 1.28m. The rock is at altitude 15 m when H(x) = 15: 15=20-13x^2 -5=-13x^2 5/13= x^2 x= +- .62 .62sec. The rock is at altitude 10 m when H(x) = 10: 10=20-13x^2 -10=-13x^2 10/13 = x^2 x= +-.88 .88sec. The rock is at 5 meter heigh when H(x) = 5: 5=20-13x^2 -15 = -13x^2 15/13=x^2 x= +- 1.07 1.07sec. To find when the rock strikes the ground let y = 0 and we get 0= 20-13x^2. Adding -20 to both sides we have -20=-13x^2. Multiplying both sides by -1/13 we get 20/13=x^2. Taking the square root of both sides we obtain the approximate value of x: x=+-1.24 We conclude that x = 1.24sec. when the rock strikes the ground **......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
22:36:55 query 3.1.90. Sales S vs. advertising expenditures A. 335 339 337 343 341 350 351 vs. 20, 22, 22.5, 24, 24, 27, 28.3 in thousands of dollars.Does the given table describe a function? Why or why not?What two points on your straight line did you pick and what is the resulting equation?What is the meaning of the slope of this line?Give your equation as a function and give the domain of the function.What is the predicted sales if the expenditures is $25,000?
......!!!!!!!!...................................
RESPONSE --> yes its a function due to increasing factors on an equal rising increasing rate 20 335 incease by 4 d=all reals 336
.................................................
......!!!!!!!!...................................
22:37:11 The table does not describe a function because ordered pairs that have the same first element and a different second element. Specifically 24,000 is paired with both 343,000 and 341,000.
I picked the points (20000,335000) (27000,350000). INSTRUCTOR COMMENT: These are data points, not points on the best-fit straight line graph. You should have sketched your line then picked two points on the line, and the line will almost never pass through data points. STUDENT SOLUTION CONTINUED: The slope between these points is slope = (350000-335000)/(27000-20000) = 15000/7000 = 15/7 = 2.143 approximately. Our equation, using this slope and the first chosen point, is therefore y-335000=2.143(x-20000) y- 335 = 2.143x-42857.143 y= 2.143x+29214.857 equation of the line Expressed as a function we have f(x) = 2.143x+292142.857. Predicted sales for expenditure $25000 will be f(25000) = 2.143(25000) + 292142.857 = 53575 + 292142.857 = 345717.857 We therefore have predicted sales f(25000)= $345,717.86 INSTRUCTOR COMMENT: Excellent solution, except for the fact that you used data points and not points on the best-fit line.......!!!!!!!!...................................
RESPONSE --> ok
.................................................
RԐv|t~ͷnj[Ј assignment #023 {g逷ẕ College Algebra 03-05-2006
......!!!!!!!!...................................
22:38:14 query 3.3.16 (was 3.2.6). Key pts and behavior: far left decr, far right incr, zeros at -10, 5, 0, 5, peaks at (-8,-4), (-2, 6), (2, 10). Local min, max among listed points.List the intervals on which the function is decreasing.
......!!!!!!!!...................................
RESPONSE --> -4 decreasin
.................................................
......!!!!!!!!...................................
22:38:22 ** The function decreases until reaching the local min at (-8, -4), then increases until reaching the local max at (-2, 6).
The function then decreases to its local min at (5, 0), after which it continues increasing. So the graph is decreasing on (-infinity, -8), on (-2, 0) and on (2, 5). **......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
22:38:35 query 3.3.22 (was 3.2.12). Piecewise linear (-3,3) to (-1,0) to (0,2) to (1,0) to (3,3).Give the intercepts of the function.Give the domain and range of the function.Give the intervals on which the function is increasing, decreasing, and constant.Tell whether the function is even, odd or neither.
......!!!!!!!!...................................
RESPONSE --> Enter, as appropriate, an answer to the question, a critiqokue of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.
Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems..................................................
......!!!!!!!!...................................
22:38:45 ** The function intersects the x axis at (-1, 0) and (1, 0), and the y axis at (0, 2).
The function decreases from (-3,3) to (-1,0) so it is decreasing on the interval (-3, -1). The function decreases from (0, 2) to (1, 0) so it is decreasing on the interval (0, 1). The function increases from (-1,0) to (0, 2) so it is increasing on the interval (-1, 0). The function ioncreases from (1,0) to (3, 3) so it is increasing on the interval (1, 3). The domain of the function is the set of possible x values, so the domain is -3 <= x <= 3, written as the interval [-3, 3]. The range of the function is the set of possible y values, so the range is 0 <= y <= 3, written as the interval [0, 3]. The function is symmetric about the y axis, with f(-x) = f(x) for every x (e.g., f(-3) = f(3) = 3; f(-1) = f(1) = 0; etc.). So the function is even. **......!!!!!!!!...................................
RESPONSE --> ok
.................................................