Assignment 26 3161

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course Mth 158

3/29/13 Around 12:00 PM

026. `*   26 

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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?

 

 

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Your solution:

 f + g = |x| + x; Domain all real numbers; Range = zero and all positive numbers

 f - g = |x| - x; Domain all real numbers; Range = zero and all positive numbers

 f * g = |x| * x; Domain all real numbers; Range all real numbers

 f / g = |x| / x; Domain {x | x ≠ 0}; Range { y | y = -1, y = 1}

 

 

confidence rating #$&*: 2

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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}. **

 

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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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?

 

 

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

 f + g = (sqrt(x + 1)*x + 2) / x; Domain {x | x >= -1, x ≠ 0}

 f - g = (sqrt(x + 1)*x - 2) / x; Domain {x | x >= -1, x ≠ 0}

 f * g = (sqrt(x + 1)*2) / x; Domain {x | x >= -1, x ≠ 0}

 f / g = (sqrt(x + 1)*x / 2; Domain {x | x >= -1, x ≠ 0}

 I'm not really sure how to figure out the ranges.

 

 

confidence rating #$&*: 2

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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. **

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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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.

 

 

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Your solution:

 g(4) = 3 / (4^2 + 2) = 3 / 18 = 1/6

 f(1/6) = |1/6 - 2| = 1*5/6 = 11/6

 

 f(2) = |2 - 2| = 0

 g(0) = 3 / (0^2 + 2) = 3/2

 

 f(1) = |1 - 2| = 1

 f(1) = |1 - 2| = 1

 

 g(0) = 3/2

 g(3/2) = 3 / ((3/2)^2 + 2) = 3 / (9/4 + 8/4) = 3 / (17/4) = 12/17

 

 

confidence 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. **

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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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?

 

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Your solution:

 {x | x >= 2} or (2 , infinity)

 

 

confidence 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}. **

 

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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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.

 

 

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Your solution:

 f(g(x)) = 2 / (2 + 3x); {x | x ≠ 0, x ≠ -2/3}

 

 g(f(x)) = (2(x + 3)) / x; {x | x ≠ 0}

 

 f(f(x)) = x / (x + 3(x + 3); {x | x ≠ -3, x ≠ -9/4}

 

 g(g(x)) = (2x) / 2; { x | x ≠ 0}

 

 

confidence rating #$&*: 2

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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. **

 

 

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Self-critique (if necessary):

 I understand what I missed. I didn't catch all of the requirements. I missed that g(f(x)) and f(g(x))'s domains can't be -3, but I understand why they aren't in those domains.

 

 

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Self-critique Rating: 3

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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.

 

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Your solution:

 f(g(x)) = g(f(x)) = x

 (x - 5) + 5 = (x + 5) - 5 = x

 x = x = x

 

 

confidence 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. **

 

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Self-critique (if necessary):

 I don't know why, but I found this problem to be quite funny.

 

 

@&

That requires a well-developed sense of humor.

*@

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Self-critique Rating: 3

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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.

 

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Your solution:

 f(x) = sqrt(x)

 g(x) = x^2 + 1

 f(g(x)) = sqrt(x^2 + 1) = H(x)

 

 

confidence 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). **

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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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?

 

 

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Your solution:

 32/81(pi)t^9

 

 

confidence 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. **

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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#*&!

&#Good responses. See my notes and let me know if you have questions. &#