Question Number 188590 by 073 last updated on 03/Mar/23
Answered by CElcedricjunior last updated on 03/Mar/23
![((f(3x))/(g(x−2)))=x^3 ;f′(3)=5;g(−1)=2 =>g′(−1)+f(3)=k trouvons k (((f(3x))/(g(x−2))))′=((3f′(3x)g(x−2)−g′(x−2)f(3x))/([g(x−2)]^2 ))=3x^2 =>((3f′(3)g(−1)−g′(−1)f(3))/((g(−1))^2 ))=3 =>((3×5×2−g′(−1)f(3))/4)=3 =>25−g′(−1)f(3)=12 ■Moivre =>g′(−1)f(3)=13 (2) ((f(3x))/(g(x−2)))=x^3 =>((f(3))/(g(−1)))=1★Cedric junior f(3)=g(−1)=>f(3)=2 (1) remplac_ξ ons (1) dans (2) =>g′(−1)×2=13=>g′(−1)=((13)/2) =>g′(−1)+f(3)=((13)/2)+2=((17)/2) =>g′(−1)+f(3)=((17)/2)](https://www.tinkutara.com/question/Q188599.png)
$$\frac{\boldsymbol{{f}}\left(\mathrm{3}\boldsymbol{{x}}\right)}{\boldsymbol{{g}}\left(\boldsymbol{{x}}−\mathrm{2}\right)}=\boldsymbol{{x}}^{\mathrm{3}} \:;\boldsymbol{{f}}'\left(\mathrm{3}\right)=\mathrm{5};\boldsymbol{{g}}\left(−\mathrm{1}\right)=\mathrm{2} \\ $$$$=>\boldsymbol{{g}}'\left(−\mathrm{1}\right)+\boldsymbol{{f}}\left(\mathrm{3}\right)=\boldsymbol{{k}}\:\boldsymbol{{trouvons}}\:\boldsymbol{{k}} \\ $$$$\left(\frac{\boldsymbol{{f}}\left(\mathrm{3}\boldsymbol{{x}}\right)}{\boldsymbol{{g}}\left(\boldsymbol{{x}}−\mathrm{2}\right)}\right)'=\frac{\mathrm{3}\boldsymbol{{f}}'\left(\mathrm{3}\boldsymbol{{x}}\right)\boldsymbol{{g}}\left(\boldsymbol{{x}}−\mathrm{2}\right)−\boldsymbol{{g}}'\left(\boldsymbol{{x}}−\mathrm{2}\right)\boldsymbol{{f}}\left(\mathrm{3}\boldsymbol{{x}}\right)}{\left[\boldsymbol{{g}}\left(\boldsymbol{{x}}−\mathrm{2}\right)\right]^{\mathrm{2}} }=\mathrm{3}\boldsymbol{{x}}^{\mathrm{2}} \\ $$$$=>\frac{\mathrm{3}\boldsymbol{{f}}'\left(\mathrm{3}\right)\boldsymbol{{g}}\left(−\mathrm{1}\right)−\boldsymbol{{g}}'\left(−\mathrm{1}\right)\boldsymbol{{f}}\left(\mathrm{3}\right)}{\left(\boldsymbol{{g}}\left(−\mathrm{1}\right)\right)^{\mathrm{2}} }=\mathrm{3} \\ $$$$=>\frac{\mathrm{3}×\mathrm{5}×\mathrm{2}−\boldsymbol{{g}}'\left(−\mathrm{1}\right)\boldsymbol{{f}}\left(\mathrm{3}\right)}{\mathrm{4}}=\mathrm{3} \\ $$$$=>\mathrm{25}−\boldsymbol{{g}}'\left(−\mathrm{1}\right)\boldsymbol{{f}}\left(\mathrm{3}\right)=\mathrm{12}\:\blacksquare{Moivre} \\ $$$$=>\boldsymbol{{g}}'\left(−\mathrm{1}\right)\boldsymbol{{f}}\left(\mathrm{3}\right)=\mathrm{13}\:\:\:\left(\mathrm{2}\right) \\ $$$$\frac{\boldsymbol{{f}}\left(\mathrm{3}\boldsymbol{{x}}\right)}{\boldsymbol{{g}}\left(\boldsymbol{{x}}−\mathrm{2}\right)}=\boldsymbol{{x}}^{\mathrm{3}} =>\frac{\boldsymbol{{f}}\left(\mathrm{3}\right)}{\boldsymbol{{g}}\left(−\mathrm{1}\right)}=\mathrm{1}\bigstar\mathscr{C}{edric}\:{junior} \\ $$$$\boldsymbol{{f}}\left(\mathrm{3}\right)=\boldsymbol{{g}}\left(−\mathrm{1}\right)=>\boldsymbol{{f}}\left(\mathrm{3}\right)=\mathrm{2}\:\left(\mathrm{1}\right) \\ $$$$\boldsymbol{{rempla}} ext{\c{c}} \boldsymbol{{cons}}\:\left(\mathrm{1}\right)\:\boldsymbol{{dans}}\:\left(\mathrm{2}\right) \\ $$$$=>\boldsymbol{{g}}'\left(−\mathrm{1}\right)×\mathrm{2}=\mathrm{13}=>\boldsymbol{{g}}'\left(−\mathrm{1}\right)=\frac{\mathrm{13}}{\mathrm{2}} \\ $$$$=>\boldsymbol{{g}}'\left(−\mathrm{1}\right)+\boldsymbol{{f}}\left(\mathrm{3}\right)=\frac{\mathrm{13}}{\mathrm{2}}+\mathrm{2}=\frac{\mathrm{17}}{\mathrm{2}} \\ $$$$=>\boldsymbol{{g}}'\left(−\mathrm{1}\right)+\boldsymbol{{f}}\left(\mathrm{3}\right)=\frac{\mathrm{17}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by qaz last updated on 03/Mar/23
$$\mathrm{3}{f}\left(\mathrm{3}{x}\right)'=\mathrm{3}{x}^{\mathrm{2}} {g}\left({x}−\mathrm{2}\right)+{x}^{\mathrm{3}} {g}\left({x}−\mathrm{2}\right)' \\ $$$$\Rightarrow\mathrm{3}{f}\left(\mathrm{3}\right)'=\mathrm{3}{g}\left(−\mathrm{1}\right)+{g}\left(−\mathrm{1}\right)'\:\:\:\Rightarrow{g}\left(−\mathrm{1}\right)'=\mathrm{9} \\ $$$$\frac{{f}\left(\mathrm{3}\right)}{{g}\left(−\mathrm{1}\right)}=\mathrm{1}\:\:\:\Rightarrow{f}\left(\mathrm{3}\right)=\mathrm{2} \\ $$$$\Rightarrow{g}\left(−\mathrm{1}\right)'+{f}\left(\mathrm{3}\right)=\mathrm{11} \\ $$
Commented by 073 last updated on 03/Mar/23
$$\mathrm{nice}\:\mathrm{solution} \\ $$
Commented by manxsol last updated on 03/Mar/23
$${that}'{s}\:{good},{simple} \\ $$
Answered by manxsol last updated on 03/Mar/23
$$\mathrm{11} \\ $$
Answered by manxsol last updated on 03/Mar/23
![((f(3x))/(g(x−2)))=x^3 .f′(3)=5 g(−1)=2 f(3)=2 E=g′ (−1)+f(3) ______________________ f(3)=(1)g(-1) f(3)=2 [f′(3x)]=3f′(3x) (I) [((f(3x))/(g(x−2)))]′=(x^3 )′ ((g(x−2)[3f′ (3x)]−g′(x−2)f(3x))/([g(x−2)]^2 ))=3x^2 x=1 3g(-1)f′ (3)−g′(-1)f(3)=3[g(−1)]^2 f′(3)=5 g(−1)=2 f(3)=2^() (3)(2)(5)−g′(-1)(2)=3(2^2 ) 30−12=2g′(-1) g′(-1)=9 E= g′(-1)+f(3) E=9+2 E=11](https://www.tinkutara.com/question/Q188616.png)
$$\frac{{f}\left(\mathrm{3}{x}\right)}{{g}\left({x}−\mathrm{2}\right)}={x}^{\mathrm{3}} \\ $$$$.{f}'\left(\mathrm{3}\right)=\mathrm{5}\:\:\:\:\:\:{g}\left(−\mathrm{1}\right)=\mathrm{2}\:\:\:{f}\left(\mathrm{3}\right)=\mathrm{2} \\ $$$${E}={g}'\:\left(−\mathrm{1}\right)+{f}\left(\mathrm{3}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${f}\left(\mathrm{3}\right)=\left(\mathrm{1}\right){g}\left(-\mathrm{1}\right) \\ $$$${f}\left(\mathrm{3}\right)=\mathrm{2} \\ $$$$\left[{f}'\left(\mathrm{3}{x}\right)\right]=\mathrm{3}{f}'\left(\mathrm{3}{x}\right)\:\:\:\:\left({I}\right) \\ $$$$\left[\frac{{f}\left(\mathrm{3}{x}\right)}{{g}\left({x}−\mathrm{2}\right)}\right]'=\left({x}^{\mathrm{3}} \right)' \\ $$$$\frac{{g}\left({x}−\mathrm{2}\right)\left[\mathrm{3}{f}'\:\left(\mathrm{3}{x}\right)\right]−{g}'\left({x}−\mathrm{2}\right){f}\left(\mathrm{3}{x}\right)}{\left[{g}\left({x}−\mathrm{2}\right)\right]^{\mathrm{2}} }=\mathrm{3}{x}^{\mathrm{2}} \\ $$$${x}=\mathrm{1} \\ $$$$\mathrm{3}{g}\left(-\mathrm{1}\right){f}'\:\left(\mathrm{3}\right)−{g}'\left(-\mathrm{1}\right){f}\left(\mathrm{3}\right)=\mathrm{3}\left[{g}\left(−\mathrm{1}\right)\right]^{\mathrm{2}} \\ $$$$\overset{} {{f}'\left(\mathrm{3}\right)=\mathrm{5}\:\:\:\:\:\:{g}\left(−\mathrm{1}\right)=\mathrm{2}\:\:\:{f}\left(\mathrm{3}\right)=\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\left(\mathrm{2}\right)\left(\mathrm{5}\right)−{g}'\left(-\mathrm{1}\right)\left(\mathrm{2}\right)=\mathrm{3}\left(\mathrm{2}^{\mathrm{2}} \right) \\ $$$$\mathrm{30}−\mathrm{12}=\mathrm{2}{g}'\left(-\mathrm{1}\right) \\ $$$${g}'\left(-\mathrm{1}\right)=\mathrm{9} \\ $$$${E}=\:{g}'\left(-\mathrm{1}\right)+{f}\left(\mathrm{3}\right) \\ $$$${E}=\mathrm{9}+\mathrm{2} \\ $$$${E}=\mathrm{11} \\ $$
Commented by 073 last updated on 03/Mar/23
$$\mathrm{thanks} \\ $$