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f and g are cont and diff functions from R→ R.
If f(x+g(y))= 4x+3y+6.
then g(2011+f(2011)) equals??

1 Answers

21
Shubhodip ·

f(x+g(0)) = 4x+ 6

let g(0) = c

f(x+ c) = 4x + 6

implied f(x) = 4(x-c) + 6 = 4x + 6 - 4c,

f(g(y)) = 3y+ 6 = 4g(y) + 6 - 4c, so g(y) = (3/4)y + c

Indeed f(x+ g(y)) = 4x+ 3y + 6

f(2011) = 8050 - 4c

2011+ f(2011) = 10061 - 4c

g(2011 + f(2011)) = (3/4)(10061 - 4c) + c , which is arbitrary...

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