course Mth 158 Mr. Smith:I will take the chapter 3 test this week. I must sent assignment 27 before this- the q/a timed out and I was probably half way through it. I apologize for the out of order sequence. Do I need to do assigmnent 27 over and resubmit. Also, I have been working with a math professor twice a week to try to finish this class. My teaching duties are killing my available time- but, That is in all reality my problem. Just wanted you to know that I am working even though the submitted assignments are slow in coming. If I pull up the q/a to work on it a little each night, will the program time out ? I could make a little quicker progresss if I could do this.
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17:20:12 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 --> a. l x l + x = (f +g) ( x) domain {x such x all real numbers} b.l x l - x domain { x such athat x is all real numbers} c.l x l *x domain { x such that x is all real numbers} d.l xl / x domain { x such that x is not 0}: ( negative infinity to 0] and [0, positive infinity). I do not think there are any restrictions on the ranges. we were not asked to find the ranges in our book and I could not find an explanation for how to address the ranges.
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17:21:19 ** 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 --> saved to my notes
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17:49:43 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 --> a. f+g sqrt rt x+1 + ( 2 / x ) domain:x >= -1 range y>= all positive real numbers b. sqrt x + 1 - ( 2/x) domain {x l >= -1 range : all real numbers c. sqrt x + 1 * 2/x ( 2* swrt x +1) / x domain: [ -1 , 0 ) ,[ 0 , all positive real numbers) range: all positive real numbers. d. sqrt x +1 / (x/2) x * sqrt x +1 / 2 domain : { x >= -1} range all positive real numbers sqrt 4 + ( 2/3) 2 2/3 domain { x l x = any real number range { yl y = all positive real numbers} f/g
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17:51:06 ** 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 --> I will ask my tutor how to determine the range of these functions.
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19:39:08 query 6.1.18 / 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 --> a. f[ g(4) ] g (x) = 3 / (x^2 +2) = 3 / ( 4^2 +2) = 1/6 b. f ( 1/6) = l (1/6) - 2 = l 11/6 l g[ f (2) ] l 2-2 l 0 g (x) = 3 /[ ( 0^2) + 2 ] = 3/2 c. f( 1) = 1 f(1) = l 1-2 l = 1 d. g [ g ( 0) ) g( 0) = 3/ 0^2 +2 = 3/2 g (3/2) = 3 / ( 3 / 2)^2+2 = 3 / ( 9/4 ) +2 = 3/ ( 17/4) = 12/17
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19:39:27 ** 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 --> ok
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19:44:08 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 --> this problem is not in 6.1 . we have done no work in chapter d yet. From looking ay the problem I think the domain is { xl x is any real number since the value os the denominator will be at least a -2.
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19:45:30 ** 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 --> I think I understand?
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19:45:44 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 --> 0k
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20:58:32 query 6.1.24 / 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) ] 2 / x / [ (2/x) ] + 3 (2 /x ) * 1/ [ 2x /x + 3x 2 / 2 + 3x domain: { x is not -2/3 and x is not 0 } g [ f (x) ] g ( x / x +3) 2 /( x / (x +3) domain: {x l x is not = to -3} f [f (x) ] f (x / x+3 ) ( x / (x +3) / ( x / (x / +3) +3 ( x / ( x +3) / ( x / (x +3) +( 3x +9 / x +3) ( x / ( x + 3) /( 4x + 9 /(x + 3 ) domain : { x l x is not equal to -3} g[ g(x)] g( 2/x) 2/ ( 2 / x) 2 * ( x/2) 2x/2 x domain { x l x = any real number}
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20:59:50 ** 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 --> I don't get it. I will show this to my tutor for clarification.
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21:03:24 query 6.1.48 / 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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21:03:36 ** f(g(x)) = g(x) + 5 = (x-5) + 5 = x.
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RESPONSE --> ok
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21:03:48 g(f(x)) = f(x) - 5 = (x+5) - 5 = x. **
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RESPONSE --> ok
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21:07:10 query 6.1.55 / 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 --> f ( x) = x g ( x) = x^2 +1 I don't understand this solution at all. Honestly, it came from the solutions manual. I could make no sense of it.
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21:07:55 ** 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 --> saved as notes
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21:09:32 query 6.1.66 / 5.1.62 (was 3.5.66). V(r) = 4 /3 pi r^3 and r(t) = 2/3 t^3, t>=0. What is the requested composite function?
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RESPONSE --> V [ r ( t) ]
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21:15:45 ** V(r(t)) = 4/3 pi * r(t)^3 = 4/3 pi * (2/3 t^3)^3 = 4/3 pi * (8/12 t^9) = 32/36 pi t^9 = 8/9 pi t^9. **
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RESPONSE --> when i tried to solve this function I got: (4 * pi / 3) *( 8/27 * t^ 9) ( 4 * pi *8 t^9)/ ( 3*27) ( 32 * pi * t^ 9)/81
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