If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution:
If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Question: * 3.1.68 / 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?
Your solution:
Confidence Assessment:
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}. **
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Question: * 3.1.72 / 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?
Your solution:
Confidence Assessment:
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. **
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Question: * 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.
Your solution:
Confidence Assessment:
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. **
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Question: * 6.1.28 / 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?
Your solution:
Confidence Assessment:
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}. **
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Question: * 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.
Your solution:
Confidence Assessment:
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. **
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Question: * 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.
Your solution:
Confidence Assessment:
Given Solution:
* * ** f(g(x)) = g(x) + 5 = (x-5) + 5 = x.
g(f(x)) = f(x) - 5 = (x+5) - 5 = x. **
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Question: * 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.
Your solution:
Confidence Assessment:
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). **
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Question: * 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?
Your solution:
Confidence Assessment:
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. **
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