Paige N.

asked • 10/16/13

Define the following functions.

1. f(x)= |x|; g(x)= |x-3|

a.) f+g
b.) f-g
c.) f⋅g
d.) f/g
e.) g/f
 
2. f(x)=1/(x+1); g(x)=x/(x-2)

a.) f+g
b.) f-g
c.) f⋅g
d.) f/g
e.) g/f
 
3. f(x)=x^2; g(x)=1/√x

a.) f+g
b.) f-g
c.) f⋅g
d.) f/g
e.) g/f

1 Expert Answer

By:

William S. answered • 10/16/13

Tutor
4.4 (10)

Experienced scientist, mathematician and instructor - William

Paige N.

Good Day Sir,
 
I am sorry for troubling you. But I am really am having a hard time to solve this equation. Can you give me the answers for each letters of Numbers 1,2,and 3? I am really in need of your answers. Once agaid, please do pardon me. :) I really will appreciate your answers.
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10/21/13

William S.

Paige, 
 
1a. abs(x)+abs(x-3) = abs(x)+abs(x-3)    No further simplification is possible.
1b. abs(x)-abs(x-3) =  abs(x)-abs(x-3)     No further simplification is possible.
1c. abs(x)*abs(x-3) =  abs [x(x-3)]           No further simplification is possible.
1d. abs(x)/abs(x-3) =  abs[x/(x-3)]          No further simplification is possible.
1e. abs(x-3)/abs(x) =  abs [(x-3)/x]         No further simplification is possible.
 
 
 
2a. [1/(x+1)] + [x/(x-2)] =  {[1/(x+1)] + [2/(x-2)] +1}  No further simplification is possible.
2b. [1/(x+1)] - [x/(x-2)] =  {[1/(x+1)] - [2/(x-2)] -1}    No further simplification is possible.
2c. [1/(x+1)] * [x/(x-2)] = {x/[(x+1)(x-2)]}                    No further simplification is possible.
2d. [1/(x+1)] / [x/(x-2)] = {(x-2)/[x(x+1)]}                     No further simplification is possible.
2e. [x/(x-2)] / [1/(x+1)] = {[x(x+1)]/(x-2)}                     No further simplification is possible
 
3a. (x^2) + (1/√x) = (x^2) + (1/√x)
3b. (x^2) - (1/√x) = (x^2) - (1/√x)
3c. (x^2) * (1/√x) = x3/2 or [(x)*(√x)]
3d. (x^2) / (1/√x) = x5/2 or [(x^2)*(√x)]
3e. (1/√x)/(x^2) = (1/x5/2) or x-5/2
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10/21/13

Paige N.

Thank you very much Sir! :) You have helped me a lot!
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10/21/13

William S.

Paige, you are very welcome.  I live to help young people understand mathematics, science, history, English and many other things.  You are the only person I've met here who thanked me, and I appreciate it very much.
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10/22/13

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