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Question Number 12246 by tawa last updated on 16/Apr/17

Let g(x) be an ininitely differentiable function , such that g(2x + 6) = g^′ (3x − 1) for all x. given that g(((44)/3)) = 33 . find g′′(8)

$$\mathrm{Let}\:\:\mathrm{g}\left(\mathrm{x}\right)\:\mathrm{be}\:\mathrm{an}\:\mathrm{ininitely}\:\mathrm{differentiable}\:\mathrm{function}\:,\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{g}\left(\mathrm{2x}\:+\:\mathrm{6}\right)\:=\:\mathrm{g}^{'} \left(\mathrm{3x}\:−\:\mathrm{1}\right)\:\mathrm{for}\:\mathrm{all}\:\mathrm{x}. \\ $$$$\mathrm{given}\:\mathrm{that}\:\:\mathrm{g}\left(\frac{\mathrm{44}}{\mathrm{3}}\right)\:=\:\mathrm{33}\:.\:\:\mathrm{find}\:\:\:\mathrm{g}''\left(\mathrm{8}\right) \\ $$

Answered by mrW1 last updated on 16/Apr/17

u=3x−1⇒x=((u+1)/3)⇒2x+6=2×((u+1)/3)+6=((2u+20)/3) g′(u)=g(((2u+20)/3)) g′′(u)=g′(((2u+20)/3))×(2/3)=(2/3)×g(((2(((2u+20)/3))+20)/3))=(2/3)×g(((4u+100)/9)) g′′(8)=(2/3)×g(((4×8+100)/9))=(2/3)×g(((44)/3))=(2/3)×33=22

$${u}=\mathrm{3}{x}−\mathrm{1}\Rightarrow{x}=\frac{{u}+\mathrm{1}}{\mathrm{3}}\Rightarrow\mathrm{2}{x}+\mathrm{6}=\mathrm{2}×\frac{{u}+\mathrm{1}}{\mathrm{3}}+\mathrm{6}=\frac{\mathrm{2}{u}+\mathrm{20}}{\mathrm{3}} \\ $$$${g}'\left({u}\right)={g}\left(\frac{\mathrm{2}{u}+\mathrm{20}}{\mathrm{3}}\right) \\ $$$${g}''\left({u}\right)={g}'\left(\frac{\mathrm{2}{u}+\mathrm{20}}{\mathrm{3}}\right)×\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}}×{g}\left(\frac{\mathrm{2}\left(\frac{\mathrm{2}{u}+\mathrm{20}}{\mathrm{3}}\right)+\mathrm{20}}{\mathrm{3}}\right)=\frac{\mathrm{2}}{\mathrm{3}}×{g}\left(\frac{\mathrm{4}{u}+\mathrm{100}}{\mathrm{9}}\right) \\ $$$${g}''\left(\mathrm{8}\right)=\frac{\mathrm{2}}{\mathrm{3}}×{g}\left(\frac{\mathrm{4}×\mathrm{8}+\mathrm{100}}{\mathrm{9}}\right)=\frac{\mathrm{2}}{\mathrm{3}}×{g}\left(\frac{\mathrm{44}}{\mathrm{3}}\right)=\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{33}=\mathrm{22} \\ $$

Commented by tawa last updated on 16/Apr/17

God bless you sir. I need the cubic equation sir. formular. i really appreciate your effort sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{need}\:\mathrm{the}\:\mathrm{cubic}\:\mathrm{equation}\:\mathrm{sir}.\:\mathrm{formular}. \\ $$$$\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}\:\mathrm{sir}. \\ $$

Commented by mrW1 last updated on 16/Apr/17

Commented by tawa last updated on 16/Apr/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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