Assignment 26

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Given Solution: * * 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}. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3

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Given Solution: * * 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3

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Given Solution: * * 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3

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Given Solution: * * 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}. 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}. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3

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Given Solution: * * 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3

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Given Solution: * * ** f(g(x)) = g(x) + 5 = (x-5) + 5 = x. g(f(x)) = f(x) - 5 = (x+5) - 5 = x. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3

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Given Solution: * * ** 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3

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Given Solution: * * ** V(r(t)) = 4/3 pi * r(t)^3 = 4/3 pi * (2/3 t^3)^3 = 4/3 pi * (8/27 t^9) = 32/81 pi t^9. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3"

Assignment 26

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&#This looks good. Let me know if you have any questions. &#