Assignment 26

course mth 158

??????yD??????assignment #026

026. `query 26

College Algebra

07-24-2008

......!!!!!!!!...................................

17:36:11

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?

......!!!!!!!!...................................

RESPONSE -->

f(x) =|x| g(x)= x

a. |x| + x Domain x|x= all real numbers

Right idea, but the grammar is wrong, and the distinction is important. x stands for only one number at a time, and the set-builder notation gives the test for whether x is a member of the set. A given x cannot be 'all real numbers'. The domain is {x | x is a real number |

b. |x| (x) Domain x|x= all real numbers

c. |x|(x) Domain x|x=all real numbers

d. |x|/x (x) Domain X|x= all real numbers

e. (|x| + x)

f. (|x| - x) (4)

g. (|x| -x) (2)

h. |x| / x (1)

.................................................

......!!!!!!!!...................................

17:39:46

** 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}. **

......!!!!!!!!...................................

RESPONSE -->

The deminominator for |x|/x can be 0 because it is not divisble by 0.

Close. It's so because division by 0 is undefined.

.................................................

......!!!!!!!!...................................

17:55:20

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?

......!!!!!!!!...................................

RESPONSE -->

a. sqrt x+1 + 2/x Domain x| x is all real numbers not equal to 0

b. sqrt x+1 - 2/x Domain x|x is all real numbers not equal to 0

c. (sqrt x +1) 2/x Domain x|x is all real numbers not equal to 0

d. sqrt x + 1/ 2/x Domain x|x is all real numbers not equal to zero

e. sqrt x +1 + 2/x (3) Domain x|x is all real numbers not equal to zero

f. sqrt x -+1 - 2/x (4)

g. sqrt x + 1 * 2/x(2)

hsqrt x+1/ 2/x (1)

.................................................

......!!!!!!!!...................................

18:00:03

** 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. **

......!!!!!!!!...................................

RESPONSE -->

So the domain for combining the function would include x< -1.

.................................................

......!!!!!!!!...................................

22:13:11

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.

......!!!!!!!!...................................

RESPONSE -->

f(g)(4)= f 3/4^2 +2 = 18

g(f)(4)= g =|x-2| = |2-2|=0

f(f)(1) = f =|1-2| = |-1|= 1

g(g(0)= 3/0^2 =2 =3/2

.................................................

......!!!!!!!!...................................

22:16:52

** 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. **

......!!!!!!!!...................................

RESPONSE -->

f(g)(4)= My fraction did not match up. I left off the numerator part and the same with the g(g) (0)

.................................................

......!!!!!!!!...................................

22:19:59

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?

......!!!!!!!!...................................

RESPONSE -->

Domain is x|x = any real numbers

.................................................

......!!!!!!!!...................................

22:21:01

** 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}.

......!!!!!!!!...................................

RESPONSE -->

Domain (-2,infinity)

.................................................

......!!!!!!!!...................................

22:21:20

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}. **

......!!!!!!!!...................................

RESPONSE -->

ok

.................................................

......!!!!!!!!...................................

22:24:58

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.

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

22:28:56

** 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. **

......!!!!!!!!...................................

RESPONSE -->

ok

.................................................

......!!!!!!!!...................................

22:41:24

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.

......!!!!!!!!...................................

RESPONSE -->

f(x+5) (x-5) = 0

g(x+5)(x-5) =0

.................................................

......!!!!!!!!...................................

22:42:40

** f(g(x)) = g(x) + 5 = (x-5) + 5 = x.

......!!!!!!!!...................................

RESPONSE -->

The -5 and +5 cancel leaving x.

.................................................

......!!!!!!!!...................................

22:43:37

g(f(x)) = f(x) - 5 = (x+5) - 5 = x. **

......!!!!!!!!...................................

RESPONSE -->

again the x-5 and x+5 cancel leaving x

.................................................

......!!!!!!!!...................................

22:45:20

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.

......!!!!!!!!...................................

RESPONSE -->

f(x) = sqrt x

g(x) =x^2 + 1

.................................................

......!!!!!!!!...................................

22:45:58

** 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). **

......!!!!!!!!...................................

RESPONSE -->

ok

.................................................

......!!!!!!!!...................................

22:48:32

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?

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

22:50:15

** 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. **

......!!!!!!!!...................................

RESPONSE -->

I don't understand this bu will work on comprehending it.

If you tell me what you do and don't understand, and self-critique the given solution, I'll be glad to try to help.

.................................................

As I believe I've noted on other assignments, more detail in self-critique would be helpful. See my notes.