[Guest]

There are two functions f(x) and g(x,y), that work in the real domain.

f(x) returns the absolute value of its input. ( e.x. f(-1) = 1 ).

g(x,y) returns numerically bigger of the two input numbers ( e.x. g(1, −2) = 1 ).

1. f(x) in terms of g(x,y)

2. g(x,y) in terms of f(x)

[Guest]

There are two functions f(x) and g(x,y), that work in the real domain.

f(x) returns the absolute value of its input. ( e.x. f(-1) = 1 ).

g(x,y) returns numerically bigger of the two input numbers ( e.x. g(1, −2) = 1 ).

1. f(x) in terms of g(x,y)

2. g(x,y) in terms of f(x)

f(x) = g(x, -x)

G(x, y) = y + f(x) + f(y)

[Guest]

f(x) can b written as

f(x) = x + 2*g(0, x)

working on the second one

[Guest]

[spoiler='for g(x, y)

']g(x,y) = f(x) - f(y) + y

[Guest]

looks suspiciously like you are trying to get us to do your homework...

[Guest]

[spoiler='for g(x, y)

']g(x,y) = f(x) - f(y) + y

i think i got it wrong, don't know about second

Edited June 4, 2009 by tarunark

[Guest]

g(x,y)= (x+y+f(x-y))/2

[Guest]

Be sure it works when both x and y are negative.

g(x,y)=x*[f(x-y)+(x-y)]/[2*(x-y)]+y*[f(y-x)+(y-x)]/[2*(y-x)]

[Guest]

Mine reduces down to mghoffman78's, so ignore mine and use the simplified one.

Edited June 4, 2009 by eel_electric

[Guest]

Be sure it works when both x and y are negative.

g(x,y)=x*[f(x-y)+(x-y)]/[2*(x-y)]+y*[f(y-x)+(y-x)]/[2*(y-x)]

If you go through the math, you will find that your solution simlifies exactly to my solution above.

EDIT: Ah, you beat me to it!

Edited June 4, 2009 by mghoffman78

[Guest]

Kudos to tkn (1) AND mghoffman78(2).. that was very fast

[bonanova]

See also.