course Mth158 I'm sorry I had so many questions on this, but I really tried my best... ³±œv¢ÕüIx’ úê}·PèQÏŸ¥¢assignment #026
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14:37:11 3.2.66 (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 --> This problem is actually 3.1.66 in the 7th edition book (a). (f+g)(x) = |x| + x, (f+g)(x) = 2x domain is x | x = the set of all real numbers range is y | y >= 0 (b). (f+g)(x) = |x| - x, (f+g)(x) = 0 domain is x | x = the set of all real numbers range is y | y >= 0 (c). (f+g)(x) = |x| * x, (f+g)(x) = x^2 domain is x | x = the set of all real numbers range is y | y >= 0 (d). (f+g)(x) = |x|/x, (f+g)(x) = 1 domain is x | x does not equal 0 range is y | y = the set of all real numbers, since if x = 1 the whole way, the graph will be a straight line infinitely up and down. That's what makes sense to me, anyways. confidence assessment: 1
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14:51:28 ** 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 --> I don't see why f*g's range was all real numbers if the graph of x^2 only includes y | y >= 0
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15:36:32 3.2.70 (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)(x) = sqrt(x+1) + 2/x I am guessing that this means there must be two graphs. domain is x | x does not equal -1 or 0 range for both graphs would be y | y >= 0 as well as y | y = all nonzero real numbers, so the combined range is going to be all real numbers. (b). (f-g)(x) = sqrt(x+1) - 2/x I am guessing again that this means there must be two graphs. domain is x | x does not equal -1 or 0 range for both graphs would be y | y >= 0 as well as y | y = all nonzero real numbers, so the combined range is going to be all real numbers. (c). (f*g)(x) = sqrt(x+1) * 2/x I have no clue whatsoever. I'll wait to enter this response and see the answer, then figure it out and see if I understand it afterwards. (d) (f/g)(x) = [sqrt(x+1)] / (2/x) Same thing. confidence assessment: 1
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15:38:54 ** 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 --> Since nothing is explained, I still don't know what to do for the last two problems (c and d) self critique assessment: 0
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16:52:46 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 --> This is actually problem 5.1.16. (a) f(g(4)) = 3/(16+2) = 1/6, f(1/6) = |1/6 - 2| = 11/6 (b) g(f(2)) = g(0) = 3/2 (c) f(f(1)) = |1-2| = f(1) = |1-2| = 1 (d) g(g(0)) = g(3/2) = 3 / (3/2 + 2) = 3 / (7/2) confidence assessment: 2
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16:54:06 ** 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 --> I see that I forgot to square 3/2 in the last one. Oops self critique assessment: 3
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17:19:34 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 --> And this is actually 5.1.26 The domain of f(g(x)) is x | x >= 2 confidence assessment: 2
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17:20:36 ** 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 don't understand why it's -2 and not 2 self critique assessment: 2
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17:21:04 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 --> Still dont understand why it's -2 and not 2 self critique assessment: 1
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18:16:13 5.2.36 (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 --> Domain of f is x | x does not equal -3 Domain of g is x | x does not equal 0 (a). f(g(x)) = f(2/x) = (2/x) / ((2/x) + 3) = (2/x) / ((2+3x) / x) = 2 / 2-x x | x equals neither 2 nor 0 (b). g(f(x)) = g(x/(x+3)) = 2 / (x / (x+3)) = 4(x+3) / x x | x equals neither 0 nor -3 (c). f(f(x)) = f(x / (x+3)) = (x/x+3) / ((x/x+3) + 3) = (x/x+3) / ((x+3(x+3) / x+3) = (x/x+3) / ((4x + 9) / x+3) = x/x+3 * (x+3)/(4x+9) = x / (4x + 3) x | x does not equal -3 (d) g(g(x)) = g(2/x) = 2 / (2/x) 2(x) / (2/x)(x) = 2x / 2 = x x | x does not equal 0 confidence assessment: 2
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18:19:46 ** 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 do not understand where you get -9/4 from or where you get -2/3 from. self critique assessment: 1
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18:24:46 5.2.46 (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 confidence assessment: 2
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18:24:56 ** f(g(x)) = g(x) + 5 = (x-5) + 5 = x.
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RESPONSE --> ok self critique assessment: 3
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18:25:02 g(f(x)) = f(x) - 5 = (x+5) - 5 = x. **
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RESPONSE --> ok confidence assessment: 3
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18:27:17 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 --> I could not find this problem anywhere, since it was not 3.5.60 in my book, and I don't have sufficient information to solve it. confidence assessment: 0
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18:27:23 ** 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 --> self critique assessment: 0
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18:42:17 5.2.62 (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 --> V(r(t)) = V((2/3)t^3) = 4/3 pi ((2/3)t^3)^3 = 4/3 pi ((8/9)t^9) = (32/27)(pi)(t^9) confidence assessment: 2
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18:43:14 ** 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 --> The equation for V in my book was V((2/3)t^3), so that's what I went by to get my answer. self critique assessment: 2
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