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Question Number 147285 by mnjuly1970 last updated on 19/Jul/21

Q: if , g (x) = 2^( ⌊x⌋ + x) , D_( g) = [0, ∞ ) then : g^( −1) (x )=? D_g^( −1) = ?

$$\:\:\:\mathrm{Q}: \\ $$$$ \\ $$$$\:{if}\:,\:\:\:{g}\:\left({x}\right)\:=\:\mathrm{2}^{\:\lfloor{x}\rfloor\:+\:{x}} \:\:,\:\mathrm{D}_{\:{g}} =\:\left[\mathrm{0},\:\infty\:\right) \\ $$$$\:\:\:\:\:\:{then}\:: \\ $$$$\:\:\:\:\:\:\:\:{g}^{\:−\mathrm{1}} \:\left({x}\:\right)=? \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{D}_{{g}^{\:−\mathrm{1}} } =\:? \\ $$$$ \\ $$

Answered by Olaf_Thorendsen last updated on 19/Jul/21

D_g = [0 ; +∞[ = ∪_(k∈N) [k ; k+1[ Let D_k = [k ; k+1[, D = ∪_(k∈N) D_k ∀x∈D_k , g(x) = 2^(⌊x⌋+x) = 2^(x+k) x+k = ((ln(g(x)))/(ln2)) = log_2 (g(x)) x = log_2 (g(x))−k ⇒ g^(−1) (x) = log_2 (x)−k g^(−1) (D_k ) = g^(−1) ([k ; k+1[) g^(−1) (D_k ) = [log_2 (k)−k ; log_2 (k+1)−k[ D_k ^(−1) = [log_2 (k)−k ; log_2 (k+1)−k[ D_g^(−1) = ∪_(k∈N) D_k ^(−1)

$$\mathrm{D}_{{g}} \:=\:\left[\mathrm{0}\:;\:+\infty\left[\:=\:\underset{{k}\in\mathbb{N}} {\cup}\left[{k}\:;\:{k}+\mathrm{1}\left[\right.\right.\right.\right. \\ $$$$\mathrm{Let}\:\mathrm{D}_{{k}} \:=\:\left[{k}\:;\:{k}+\mathrm{1}\left[,\:\mathrm{D}\:=\:\underset{{k}\in\mathbb{N}} {\cup}\mathrm{D}_{{k}} \right.\right. \\ $$$$\forall{x}\in\mathrm{D}_{{k}} ,\:{g}\left({x}\right)\:=\:\mathrm{2}^{\lfloor{x}\rfloor+{x}} \:=\:\mathrm{2}^{{x}+{k}} \\ $$$${x}+{k}\:=\:\frac{\mathrm{ln}\left({g}\left({x}\right)\right)}{\mathrm{ln2}}\:=\:\mathrm{log}_{\mathrm{2}} \left({g}\left({x}\right)\right) \\ $$$${x}\:=\:\mathrm{log}_{\mathrm{2}} \left({g}\left({x}\right)\right)−{k} \\ $$$$\Rightarrow\:{g}^{−\mathrm{1}} \left({x}\right)\:=\:\mathrm{log}_{\mathrm{2}} \left({x}\right)−{k} \\ $$$$\:{g}^{−\mathrm{1}} \left(\mathrm{D}_{{k}} \right)\:=\:{g}^{−\mathrm{1}} \left(\left[{k}\:;\:{k}+\mathrm{1}\left[\right)\right.\right. \\ $$$$\:{g}^{−\mathrm{1}} \left(\mathrm{D}_{{k}} \right)\:=\:\left[\mathrm{log}_{\mathrm{2}} \left({k}\right)−{k}\:;\:\mathrm{log}_{\mathrm{2}} \left({k}+\mathrm{1}\right)−{k}\left[\right.\right. \\ $$$$\:\mathrm{D}_{{k}} ^{−\mathrm{1}} \:=\:\left[\mathrm{log}_{\mathrm{2}} \left({k}\right)−{k}\:;\:\mathrm{log}_{\mathrm{2}} \left({k}+\mathrm{1}\right)−{k}\left[\right.\right. \\ $$$$\mathrm{D}_{{g}^{−\mathrm{1}} } \:=\:\underset{{k}\in\mathbb{N}} {\cup}\mathrm{D}_{{k}} ^{−\mathrm{1}} \\ $$

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