course mth 158
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11:25:37 query 3.4.14 (was 3.3.6). Concave down then concave up. Does this graph represent a constant, linear, square, cube, square root, reciprocal, abs value or greatest integer function?
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RESPONSE --> this graph represents a cube function
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11:25:41 ** 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. **
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RESPONSE --> ok
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22:29:41 query 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?
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RESPONSE --> the sketch increases and at an increasing rate
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22:30:43 ** 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. **
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RESPONSE --> see where I went wrong I was half right.
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22:33:41 What three points did you label on your graph?
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RESPONSE --> (1,1),(2,1.414), and (5,2.236)
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22:49:55 query 3.4.24 (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
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RESPONSE --> you mean query 3.4. 34 intercepts (0,5) and (-3,0) domain {xlx>or = to -3} range {yl<=to -3 y <= 5}
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22:52:23 ** 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). **
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RESPONSE --> I was way off but understand now
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۾z}r~jzj}YWͨy assignment #025 ؾ{]Uӂ͛ξ College Algebra 11-19-2005
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22:58:28 query 3.5.12 (was 3.4.6). Upward parabola vertex (0, 2). What equation matches this function?
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RESPONSE --> g(x)=x^2+2 maths this function
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23:03:09 The correct equation is y = x^2 + 2. How can you tell it isn't y = 2 x^2 + 2?. How can you tell it isn't y = (x+2)^2.
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RESPONSE --> because y=(x+2)^2 would be a horizontal shift of y=x^2 by looking at the points we know that it was a vertical shift.and y=2x^2+2 would be a vertical stretch not a vertical shift.
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23:03:35 GOOD STUDENT ANSWERS: it isnt y = 2x^2 + 2 because that would make it a verical stretch and it isnt vertical it is just a parabola. it isnt y = (x+2)^2 because that would make it shift to the left two units and it did not. INSTRUCTOR NOTE: Good answers. Here is more detail: The basic y = x^2 parabola has its vertex at (0, 0) and also passes through (-1, 1) and (1, 1). y = 2 x^2 would stretch the basic y = x^2 parabola vertically by factor 2, which would move every point twice as far from the x axis. This would leave the vertex at (0, 0) unchanged and would move the points (-1, 1) and (1, 1) to (-1, 2) and (1, 2). Then y = 2x^2 + 2 would shift this function vertically 2 units, so that the points (0, 0), (-1, 2) and (1, 2) would shift 2 units upward to (0, 2), (-1, 4) and (1, 4). The resulting graph coincides with the given graph at the vertex but because of the vertical stretch it is too steep to match the given graph. The function y = (x + 2)^2 shifts the basic y = x^2 parabola 2 units to the left so that the vertex would move to (-2, 0) and the points (-1, 1) and (1, 1) to (-3, 1) and (-1, 1). The resulting graph does not coincide with the given graph. y = x^2 + 2 would shift the basic y = x^2 parabola vertically 2 units, so that the points (0, 0), (-1, 1) and (1, 1) would shift 2 units upward to (0, 2), (-1, 3) and (1, 3). These points do coincide with points on the given graph, so it is the y = x^2 + 2 function that we are looking for. **
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RESPONSE --> o.k.
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23:09:03 query 3.5.16 (was 3.4.10). Downward parabola.
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RESPONSE --> It would bey= -x^2
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23:10:17 The correct equation is y = -2 x^2. How can you tell it isn't y = 2 x^2?. How can you tell it isn't y = - x^2?
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RESPONSE --> I don't understand it looks to me like it point of origin is (0,0) not - 2 .
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23:10:57 ** The basic y = x^2 parabola has its vertex at (0, 0) and also passes through (-1, 1) and (1, 1). y = -x^2 would stretch the basic y = x^2 parabola vertically by factor -1, which would move every point to another point which is the same distance from the x axis but on the other side. This would leave the vertex at (0, 0) unchanged and would move the points (-1, 1) and (1, 1) to (-1, -1) and (1, -1). y = -2 x^2 would stretch the basic y = x^2 parabola vertically by factor -2, which would move every point to another point which is twice the distance from the x axis but on the other side. This would leave the vertex at (0, 0) unchanged and would move the points (-1, 1) and (1, 1) to (-1, -2) and (1, -2). This graph coincides with the given graph. y = 2 x^2 would stretch the basic y = x^2 parabola vertically by factor 2, which would move every point twice the distance from the x axis and on the same side. This would leave the vertex at (0, 0) unchanged and would move the points (-1, 1) and (1, 1) to (-1, 2) and (1, 2). This graph is on the wrong side of the x axis from the given graph. **
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RESPONSE --> o.k.
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23:13:42 query 3.5.18 (was 3.4.12). V with vertex at origin. What equation matches this function?
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RESPONSE --> y=2lxl
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23:19:35 The correct equation is y = 2 | x | . How can you tell it isn't y = | x | ?. How can you tell it isn't y = | x | + 2?
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RESPONSE --> it isn't y= l x l because if you substitute points in the equation they do not match graph. it is not y=l x l + 2 because that would shift the graph to the left 2 units.
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23:19:59 ** The basic y = | x | graph is in the shape of a V with a 'vertex' at (0, 0) and also passes through (-1, 1) and (1, 1). The given graph has the point of the V at the origin but passes above (-1, 1) and (1, 1). y = | x | + 2 would shift the basic y = | x | graph vertically 2 units, so that the points (0, 0), (-1, 1) and (1, 1) would shift 2 units upward to (0, 2), (-1, 3) and (1, 3). The point of the V would lie at (0, 2). None of these points coincide with points on the given graph y = 2 | x | would stretch the basic y = | x | graph vertically by factor 2, which would move every point twice the distance from the x axis and on the same side. This would leave the point of the 'V' at (0, 0) unchanged and would move the points (-1, 1) and (1, 1) to (-1, 2) and (1, 2). This graph coincides with the given graph. **
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RESPONSE --> o.k.
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23:33:02 query 3.5.30 (was 3.4.24). Transformations on y = sqrt(x). What basic function did you start with and, in order, what transformations were required to obtain the graph of the given function?
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RESPONSE --> it is a square root function we then had to shift up 2 units y=sqrtx+2 we then need to reflect the y axis so we get y=sqrt-x+2 we then need to shift left 3 units we get y= sqrt3-x +2
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23:33:31 What is the function after you shift the graph up 2 units?
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RESPONSE --> y=sqrtx +2
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23:34:02 ERRONEOUS STUDENT RESPONSE: y = x^2 + 2 INSTRUCTOR CORRECTION: y = sqrt(x) is not the same as y = x^2. y = sqrt(x) is the square root of x, not the square of x. Shifting the graph of y = sqrt(x) + 2 up so units we would obtain the graph y = sqrt(x) + 2. **
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RESPONSE --> o.k.
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23:34:43 What is the function after you then reflect the graph about the y axis?
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RESPONSE --> y=sqrt(-x) +2
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23:34:57 ** To reflect a graph about the y axis we replace x with -x. It is the y = sqrt(x) + 2 function that is being reflected so the function becomes y = sqrt(-x) + 2. **
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RESPONSE --> o.k.
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23:35:36 What is the function after you then fhist the graph left 3 units?
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RESPONSE --> y=sqrt(3-x) +2
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23:36:57 ** To shift a graph 3 units to the left we replace x with x + 3. It is the y = sqrt(-x) + 2 function that is being reflected so the function becomes y = sqrt( -(x+3) ) + 2. **
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RESPONSE --> o.k i wasn't sure how to write it using this program I but that aside i had the right response.
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23:52:57 query 3.5.42 (was 3.4.36). f(x) = (x+2)^3 - 3. What basic function did you start with and, in order, what transformations were required to obtain the graph of the given function? Describe your graph. Give three points on your graph and tell to which basic points on the graph of your basic function each corresponds.
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RESPONSE --> basic function y= x^3 we then shift horizontallyto the right 2 unitswich gets us y=(x+2)^3 we the vertically shift down 3 units y=(x+2)^3 - 3 It first has a convex the a concave runs from negutive to positive. my 3 points (0,5) (1,24) (-2,0) (0,5)=(0,0) in the basic (1,24)=(1, 1) in the basic (-2,0)=(-2,-8) in the basic
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23:53:09 ** Starting with y = x^3 we replace x by x - 3 to shift the graph 3 units left. We obtain y = (x + 3)^3. We then shift this graph 2 units vertically by adding 2 to the value of the function, obtaining y = (x + 3 )^3 + 2. The basic points of the y = x^3 graph are (-1, -1), (0, 0) and (1, 1). Shifted 3 units left and 2 units up these points become (-1 - 3, -1 + 2) = (-4, 1), (0 - 3, 0 + 2) = (-3, 2) and (1 - 3, 1 + 2) = (-2, 3). The points you give might not be the same three points but should be obtained by a similar process, and you should specify both the original point and the point to which it is transformed. **
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RESPONSE --> o.k.
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00:22:27 query 3.5.58 (was 3.4.40). h(x) = 4 / x + 2. What basic function did you start with and, in order, what transformations were required to obtain the graph of the given function? Describe your graph. Give three points on your graph and tell to which basic points on the graph of your basic function each corresponds.
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RESPONSE --> basic function y=1/x or reciprocal function we then vertically stretch it 4 units it becomes y= 4/x we then horizantally shift it tho the left 2 units it becomes. y= 4/x+2 it is 2 curved lines each the recipical of the other never touching. (-1, -2) (1,6) (2,4) (-1,-2)= (-1,-1) in the basic function (1,6)= (1,1) in the basic function (2,4)= (2, .5)
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00:22:46 ** We start with the basic reciprocal function y = 1 / x, which has vertical asymptote at the y axis and horizontal asymptotes at the right and left along the x axis and passes through the points (-1, -1) and (1, 1). To get y = 4 / x we must multiply y = 1 / x by 4. Multiplying a function by 4 results in a vertical stretch by factor 4, moving every point 4 times as far from the x axis. This will not affect the location of the vertical or horizontal asymptotes but for example the points (-1, -1) and (1, 1) will be transformed to (-1, -4) and (1, 4). At every point the new graph will at this point be 4 times as far from the x axis. At this point we have the graph of y = 4 / x. The function we wish to obtain is y = 4 / x + 2. Adding 2 in this mannerincreases the y value of each point by 2. The point (-1, -4) will therefore become (-1, -4 + 2) = (-1, -2). The point (1, 4) will similarly be raised to (1, 6). Since every point is raised vertically by 2 units, with no horizontal shift, the y axis will remain an asymptote. As the y = 4 / x graph approaches the x axis, where y = 0, the y = 4 / x + 2 graph will approach the line y = 0 + 2 = 2, so the horizontal asymptotes to the right and left will consist of the line y = 2. Our final graph will have asymptotes at the y axis and at the line y = 2, with basic points (-1, -2) and (1, 6).
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RESPONSE --> o.k.
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~F|클HͻgɍUż assignment #026 ؾ{]Uӂ͛ξ College Algebra 11-22-2005
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20:25:02 query 3.1.66 (was 3.5.6). f+g, f-g, f*g and f / g for | x | and x. What are f+g, f-g, f*g and f / g and what is the domain and range of each?
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RESPONSE --> the domain for all of these equations are the same any number x that is in the domain of f and g and the range is difference of between f and g
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20:25:17 ** The domain of f is all real numbers and its range is all positive numbers. The domain of g is all real numbers and its range is all real numbers. We recall that if x < 0 it follows that | x | = -x, whereas for x > 0 we have | x | = x. The domain of f + g is all real numbers. f + g = | x | + x. Since for negative x we have | x | = -x, when x < 0 the value of f + g is zero. For x = 0 we have f + g = 0 and for x > 0 we have f + g > 0, and f + g can take any positive value. More specifically for positive x we have f + g = 2x, and for positive x 2x can take on any positive value. The range of f + g is therefore all non-negative real numbers. The domain of f - g is all real numbers. f - g = | x | - x. Since for positive x we have | x | = x, when x > 0 the value of f - g is zero. For x = 0 we have f + g = 0 and for x < 0 we have f - g > 0, and f + g can take any positive value. More specifically for negative x we have f - g = -2x, and for negative x the expression -2x can take on any positive value. The range of f - g is therefore all non-negative numbers. The domain of f * g is all real numbers. f * g = | x | * x. For x < 0 then f * g = -x * x = -x^2, which can take on any negative value. For x = 0 we have f * g = 0 and for x > 0 we have f * g = x^2, which can take on any positive value. The range of f * g is therefore all real numbers. The domain of f / g = | x | / x is all real numbers for which the denominator g is not zero. Since g = 0 when x = 0 and only for x = 0, the domain consists of all real numbers except 0. For x < 0 we have | x | / x = -x / x = -1 and for x > 0 we have | x | / x = x / x = 1. So the range of f / g consists of just the value 1 and -1; we express this as the set {-1, 1}. **
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RESPONSE --> o.k.
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20:46:04 query 3.1.70 (was 3.5.10). f+g, f-g, f*g and f / g for sqrt(x+1) and 2/x. What are f+g, f-g, f*g and f / g and what is the domain and range of each?
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RESPONSE --> sqrt(x+1)+2/x domain = {xlx not=to 0} sqrt(x+1)-2/x domain= {xlx not =to 0} sqrt(x+1)* 2/x domain= {xlx not=to 0} sqrt(x+1)/(2/x) domain= {xlx not=to 0} range for all is the highest and lowest real number for bothe f and g
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20:46:31 ** The square root is always positive and the argument of the square root must be nonnegative, so sqrt(x + 1) is defined only when x+1 > 0 or x > -1. So the domain of f is all real numbers greater than or equal to -1 and its range is all positive numbers. The function g(x) = 2/x is defined for all values of x except 0, and 2/x = y means that x = 2 / y, which gives a value of x for any y except 0. So the domain of g is all real numbers except 0 and its range is all real numbers except 0. Any function obtained by combining f and g is restricted at least to a domain which works for both functions, so the domain of any combination of these functions excludes values of x which are less than -1 and x = 0. The domain will therefore be at most {-1,0) U (0, infinity). Other considerations might further restrict the domains. The domain of f + g is {-1,0) U (0, infinity). There is no further restriction on the domain. The domain of f - g is {-1,0) U (0, infinity). There is no further restriction on the domain. The domain of f * g is {-1,0) U (0, infinity). There is no further restriction on the domain. The domain of f / g = | x | / x is {-1,0) U (0, infinity) for which the denominator g is not zero. Since the denominator function g(x) = 2/x cannot be zero there is no further restriction on the domain. **
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RESPONSE --> o.k.
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21:20:01 query 5.1.16 (was 3.5.20?). f(g(4)), g(f(2)), f(f(1)), g(g(0)) for |x-2| and 3/(x^2+2) Give the requested values in order and explain how you got each.
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RESPONSE --> f(g(4))=[3/((4-2)^2+2)]=f[3/6]=l1/2l=1/2 g(f(2))=g[l2-2l]=g(2-2)=3/((2-2)^2+2=g(2-2)=3/(0^2+2)= 3/2 f(f(1))=f(l1-2l)= f(-1)=l-1l=1 g(g(0))=g(3/(0^2+2)=g(3/2)=3/((3/2)^2+2)=12/17
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21:20:35 ** f(g(4)) = | g(4) - 2 | = | 3 / (4^2 + 2) - 2 | = | 3/18 - 2 | = | 1/6 - 12/6 | = | -11/6 | = 11/6. g(f(2)) = 3 / (f(2)^2 + 2) = 3 / ( | 2-2 | ) ^2 + 2) 3 / (0 + 2) = 3/2. f(f(1)) = | f(1) - 2 | = | |1-2| - 2 | = | |-1 | - 2 | = | 1 - 2 | = |-1| = 1. g(g(0)) = 3 / (g(0)^2 + 2) = 3 / ( (3 / ((0^2+2)^2) ^2 + 2)) = 3 / (9/4 + 2) = 3/(17/4) = 12/17. **
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RESPONSE --> o.k.
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21:27:30 query 5.2.16 (was 3.5.30). Domain of f(g(x)) for x^2+4 and sqrt(x-2) What is the domain of the composite function?
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RESPONSE --> domain = {xlxnot = to 2 not=to -2 not =to sqrt2}
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21:28:02 ** The domain of g(x) consists of all real numbers for which x-2 >= 0, i.e., for x >= -2. The domain is expressed as {-2, infinity}.
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RESPONSE --> not what i got but i understan why.
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21:28:36 The domain of f(x) consists of all real numbers, since any real number can be squared and 4 added to the result. The domain of f(g(x)) is therefore restricted only by the requirement for g(x) and the domain is {-2, infinity}. **
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RESPONSE --> o.k.
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22:08:58 query 5.1.26 (was 3.5.40). f(g(x)), g(f(x)), f(f(x)), g(g(x)) for x/(x+3) and 2/x Give the four composites in the order requested and state the domain for each.
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RESPONSE --> f(g(x))=f(2/x)=2/x/((2/x)+3)=2/2+3=2/5 domain{xlx>= 2/5} g(f(x))=g(x/x+3)=2/x/x+3=2(x+3)/x domain= {xlxnot=0,x>1} f(f(x))=f(x/x+3)=x/x+3/(x/x+3)+3=x/x+3/2x+3/x+3= x/2x+3 domain={xlx>=0} g(g(x))=g(2/x)=2/(2/x)=2x/2=x domain={xlxnot=0}
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22:09:35 ** The domain of f(x) is all x except -3. The domain of g(x) is all x except 0. The domain of f(g(x)) consists of all x for which the argument of g is not zero and for which the argument of f is not -3. The argument of g is x so x cannot be zero and the argument of f is g(x) so g(x) cannot be -3. This means that 2/x = -3 is not possible. Solving this for x we find that x cannot be -2/3. The domain of f(g(x)) is therefore all real numbers except -3 and -2/3. The domain of f(f(x)) consists of all x for which the argument of the first f is not -3 and for which the argument of the second f is not -3. The argument of the second f is x so x cannot be -3 and the argument of the first f is f(x) so f(x) cannot be -3. This means that x/(x+3) = -3 is not possible. Solving this for x we find that x cannot be -9/4. The domain of f(f(x)) is therefore all real numbers except -3 and -9/4. The domain of g(f(x)) consists of all x for which the argument of f is not -3 and for which the argument of g is not 0. The argument of f is x so x cannot be -3 and the argument of g is f(x) so f(x) cannot be 0. f(x) is zero if and only if x = 0. The domain of g(f(x)) is therefore all real numbers except -3 and 0. The domain of g(g(x)) consists of all x for which the argument of the first g is not 0 and for which the argument of the second g is not 0. The argument of the second g is x so x cannot be 0 and the argument of the first g is g(x) so g(x) cannot be 0. There is no real number for which g(x) = 2/x is zero. The domain of g(g(x)) is therefore all real numbers except 0. **
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RESPONSE --> o.k.
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22:15:35 query 5.1.46 (was 3.5.50). f(g(x)) = g(f(x)) = x for x+5 and x-5 Show f(g(x)) = g(f(x)) = x for the given functions.
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RESPONSE --> f(g(x))=f(x-5)=x+5-5=x g(f(x))=g(x+5)=x-5+5=x
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22:15:48 ** f(g(x)) = g(x) + 5 = (x-5) + 5 = x.
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RESPONSE --> o.k.
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22:15:58 g(f(x)) = f(x) - 5 = (x+5) - 5 = x. **
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RESPONSE --> o.k.
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22:18:44 query 5.1.53 (was 3.5.60). H(x) = sqrt(x^2 + 1) = f(g(x)) Give the functions f and g such that H is the composite.
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RESPONSE --> H(x) = sqrt(x^2 + 1) f(x)=sqrtx and g(x)=x^2+1
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22:19:11 ** The composite f(g(x)) has 'innermost' function g(x), to which the f function is applied. The 'innermost' function of sqrt(x^2 + 1) is x^2 + 1. The square root is applied to this result. So H(x) = f(g(x)) with f(u) = sqrt(u) and g(x) = x^2 + 1. Thus f(g(x)) = sqrt(g(x)) = sqrt(x^2 + 1). **
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RESPONSE --> o.k.
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22:22:30 query 5.1.62 (was 3.5.66). V(r) = 4/3 pi r^2 and r(t) = 2/3 t^3, t>=0. What is the requested composite function?
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RESPONSE --> r(t)=2/3t^3
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22:23:12 ** V(r(t)) = 4/3 pi * r(t)^2 = 4/3 pi * (2/3 t^3)^2 = 4/3 pi * (4/9 t^6) = 16/27 pi t^6. **
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RESPONSE --> did notr understand what you was asking for but i understand now.
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Y՜[͋ۗr assignment #027 ؾ{]Uӂ͛ξ College Algebra 11-26-2005
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22:38:25 query 3.6.6. x = -20 p + 500, 0<=p<=25 What is the revenue function and what is the revenue if 20 units are sold?
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RESPONSE --> revenue function = x(-20x+500)=-20x^2+500x what is the revenue if 20 units are sold? R(20)=-20(20)^2+500(20)=-8000+10000=2000 $2000.00
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22:40:14 ** revenue = demand * price = x * p = (-20 p + 500) * p = -20 p^2 + 500 p If price = 24 then we get R = -20 * 24^2 + 500 * 24 = 480. **
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RESPONSE --> the book say if 20 units not 24 where sold so why did you use 24 i'm confused.
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22:57:56 query 3.6.10. P = (x, y) on y = x^2 - 8. Give your expression for the distance d from P to (0, -1)
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RESPONSE --> d=sqrt(x-0)^2+(y-1)^2=sqrtx^2+y^2-1
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22:59:30 ** P = (x, y) is of the form (x, x^2 - 8). So the distance from P to (0, -1) is sqrt( (0 - x)^2 + (-1 - (x^2-8))^2) = sqrt(x^2 + (-7-x^2)^2) = sqrt( x^2 + 49 - 14 x^2 + x^4) = sqrt( x^4 - 13 x^2 + 49). **
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RESPONSE --> see where i went wrong
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23:03:06 What are the values of d for x=0 and x = -1?
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RESPONSE --> d(0)=sqrt8 d(1)sqrt1-1+8=sqrt8
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23:03:27 ** If x = 0 we have sqrt( x^4 - 13 x^2 + 49) = sqrt(0^4 - 13 * 0 + 49) = sqrt(49) = 7. If x = -1 we have sqrt( x^4 - 13 x^2 + 49) = sqrt((-1)^4 - 13 * (-1) + 49) = sqrt( 64) = 8. Note that these results are the distances from the x = 0 and x = 1 points of the graph of y = x^2 - 8 to the point (0, -1). You should have a sketch of the function and you should vertify that these distances make sense. **
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RESPONSE --> o.k.
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23:09:46 query 3.6. 18 (was and remains 3.6.18). Circle inscribed in square. What is the expression for area A as a function of the radius r of the circle?
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RESPONSE --> area of square would be 2 times the diameterof the circle so A=2D
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23:10:31 ** A circle inscribed in a square touches the square at the midpoint of each of the square's edges; the circle is inside the square and its center coincides with the center of the square. A diameter of the circle is equal in length to the side of the square. If the circle has radius r then the square has sides of length 2 r and its area is (2r)^2 = 4 r^2. The area of the circle is pi r^2. So the area of the square which is not covered by the circle is 4 r^2 - pi r^2 = (4 - pi) r^2. **
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RESPONSE --> same as 2times diameter o.k.
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23:11:24 What is the expression for perimeter p as a function of the radius r of the circle?
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RESPONSE --> p=8r
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23:11:37 ** The perimeter of the square is 4 times the length of a side which is 4 * 2r = 8r. **
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RESPONSE --> o.k.
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23:18:35 query 3.6.27 (was 3.6.30). one car 2 miles south of intersection at 30 mph, other 3 miles east at 40 mph Give your expression for the distance d between the cars as a function of time.
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RESPONSE --> d^2=(2-30t)^2+(3-40t)^2 d(t)=sqrt(2-30t)^2+(3-40t)^2 d(t)=sqrt2500t^2-360t+13
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23:18:56 ** At time t the position of one car is 2 miles south, increasing at 30 mph, so its position function is 2 + 30 t. The position function of the other is 3 + 40 t. If these are the x and the y coordinates of the position then the distance between the cars is distance = sqrt(x^2 + y^2) = sqrt( (2 + 30 t)^2 + (3 + 40t)^2 ) = sqrt( 4 + 120 t + 900 t^2 + 9 + 240 t + 1600 t^2) = sqrt( 2500 t^2 + 360 t + 13). **
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RESPONSE --> o.k.
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{ܺp֮ރ} assignment #027 ؾ{]Uӂ͛ξ College Algebra 11-26-2005 lSLD~ۚr assignment #028 ؾ{]Uӂ͛ξ College Algebra 11-26-2005
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20:00:15 4.1.42 (was 4.1.30). Describe the graph of f(x)= x^2-2x-3 as instructed.
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RESPONSE --> f(x)= x^2-2x-3 a=1,b=-2,c=-3 since a=1>0 the graph opens up the x coordinate of the vertex is x=-b/2a=--2/2=2/2=1 the y coordinate of the vertex is 1^2-2*1-3=1-2-3=-4 vertex is (1,-4) axis of symmetry is line x=1 discriminate 4-(-12)=16 16>0 so it has 2 x intercepts x intercept x^2-2x-3=0 (x+1)(x-3)=xintercept is -1 & 3 y intercept is -3 domain =all real numbers range={y|y>=-4} f is increasing on (1,infinity) f is decreasing on(-infinity, 1)
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20:03:16 The function is of the form y = a x^2 + b x + c with a = 1, b = -2 and c = -3. The graph of this quadratic function will open upwards, since a > 0. The axis of symmetry is the line x = -b / (2 a) = -(-2) / (2 * 1) = 1. The vertex is on this line at y = f(1) = 1^2 - 2 * 1 - 3 = -4. So the the vertex is at the point (1, -4). The x intercepts occur where f(x) = x^2 - 2 x - 3 = 0. We can find the values of x by either factoring or by using the quadratic formula. Here will will factor to get (x - 3) ( x + 1) = 0 so that x - 3 = 0 OR x + 1 = 0, giving us x = 3 OR x = -1. So the x intercepts are (-1, 0) and (3, 0). The y intercept occurs when x = 0, giving us y = f(0) = -3. The y intercept is the point (0, -3).
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RESPONSE --> is this the domain and range and increasing and decreasing that i had solved for it did not answer these in the solution given. i think i done excellant on that one domain =all real numbers range={y|y>=-4} f is increasing on (1,infinity) f is decreasing on(-infinity, 1)
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20:35:02 4.1.67 (was 4.1.50). What are your quadratic functions whose x intercepts are -5 and 3, for the values a=1; a=2; a=-2; a=5?
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RESPONSE --> problem in book used the x intercepts of -3and 1so f(x) =1(x+3)(x-1)=1(x^2+2x-3)=x^2+2x-3 f(x) =2(x+3)(x-1)=2(x^2+2x-3)=2x^2+4x-6 f(x) =-2(x+3)(x-1)=-2(x^2+2x-3)=-2x^2-4x+6 f(x) =5(x+3)(x-1)=5(x^2+2x-3)=5x^2+10x-15 the x intercept value is not changed the y intercept value is changed bythe diff. values of a=1,2,-2,5 the axis of symmetry is unaffected by the value of a the y part of the vertex(x,y) is mulitplied by a
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20:35:47 Since (x+5) = 0 when x = -5 and (x - 3) = 0 when x = 3, the quadratic function will be a multiple of (x+5)(x-3). If a = 1 the function is 1(x+5)(x-3) = x^2 + 2 x - 15. If a = 2 the function is 2(x+5)(x-3) = 2 x^2 + 4 x - 30. If a = -2 the function is -2(x+5)(x-3) = -2 x^2 -4 x + 30. If a = 5 the function is 5(x+5)(x-3) = 5 x^2 + 10 x - 75.
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RESPONSE --> ok
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20:36:51 Does the value of a affect the location of the vertex?
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RESPONSE --> the y coordinate of the vertex is multiplied by a
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20:37:38 In every case the vertex, which lies at -b / (2a), will be on the line x = -1. The value of a does not affect the x coordinate of the vertex, which lies halfway between the zeros at (-5 + 3) / 2 = -1.
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RESPONSE --> ok
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20:38:26 The y coordinate of the vertex does depend on a. Substituting x = -1 we obtain the following: For a = 1, we get 1 ( 1 + 5) ( 1 - 3) = -12. For a = 2, we get 2 ( 1 + 5) ( 1 - 3) = -24. For a = -2, we get -2 ( 1 + 5) ( 1 - 3) = 24. For a = 5, we get 5 ( 1 + 5) ( 1 - 3) = -60. So the vertices are (-1, -12), (-1, -24), (-1, 24) and (-1, -60).
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RESPONSE --> ok
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21:02:27 4.1.78 (was 4.1.60). A farmer with 2000 meters of fencing wants to enclose a rectangular plot that borders on a straight highway. If the farmer does not fence the side along the highway, what is the largest area that can be enclosed?
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RESPONSE --> a(x)=(2000-2x)x=2000x-2x^2=-2x^2+2000x x=-b/2a=-2000/2*-2=500 a(500)=-2(500)^2+2000(500) -500000+1000000= 500,000sq meters is the maximum area that can be enclosed
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21:02:43 ** If the farmer uses 2000 meters of fence, and fences x feet along the highway, then he have 2000 - x meters for the other two sides. So the dimensions of the rectangle are x meters by (2000 - x) / 2 meters. The area is therefore x * (2000 - x) / 2 = -x^2 / 2 + 1000 x. The graph of this function forms a downward-opening parabola with vertex at x = -b / (2 a) = -1000 / (2 * -1/2) = 1000. At x = 1000 the area is -x^2 / 2 + 1000 x = -1000^2 / 2 + 1000 * 1000 = 500,000, meaning 500,000 square meters. Since this is the 'highest' point of the area vs. dimension parabola, this is the maximum possible area. **
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RESPONSE --> ok
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21:24:20 4.1.101 (was 4.1.80). A rectangle has one vertex on the line y=10-x, x>0, another at the origin, one on the positive x-axis, and one on the positive y-axis. Find the largest area that can be enclosed by the rectangle.
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RESPONSE --> y=10-x A(x)=x(10-x)=-x^2+10x maximum is x=b/2a=-10/2*-1=10/2=5 plug 5 in for x a(5)=-x^2+10x=-25+50=25the largest area that can be enclosed is 25
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21:24:26 ** The dimensions of the rectangle are x and y = 10 - x. So the area is area = x ( 10 - x) = -x^2 + 10 x. The vertex of this rectangle is at x = -b / (2 a) = -10 / (2 * -1) = 5. Since the parabola opens downward this value of x results in a maximum area, which is -x^2 + 10 x = -5^2 + 10 * 5 = 25. **
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RESPONSE --> ok
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