enjoy this :)

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

1 Answers

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...

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω