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Define the following functions.

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1 Answer

1a. f+g = abs (x) + abs (x-3)
1b. f-g = abs (x) - abs (x-3)
1c. f*g = abs [x(x-3)]
1d. f/g = abs[x/(x-3)]
1e. g/f = abs [(x-3)/x]
 
2a. f+g = {[1/(x+1)] + [2/(x-2)] +1}
2b. f-g = {[1/(x+1)] -[2/(x-2)] -1}
2c. f*g = {x/[(x+1)(x-2)]}
2d. f/g = {(x-2)/[x(x+1)]}
2e. g/f = {[x(x+1)]/(x-2)}
 
3a. f+g = x2 + [1/(x½)]3b. 
3b f-g = x2 - [1/(x½)]
3c. f*g = x3/2
3d. f/g = x5/2
3e. = 1(x5/2)

Comments

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.
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
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.