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Question Number 153972 by SANOGO last updated on 12/Sep/21
soit:f→x^3 +3x+1 alors:(f^(_1) )^(′′) (5)=?
$${soit}:{f}\rightarrow{x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{1} \\ $$$${alors}:\left({f}^{\_\mathrm{1}} \right)^{''} \left(\mathrm{5}\right)=? \\ $$
Answered by mr W last updated on 12/Sep/21
f(x)=y=x^3 +3x+1 x^3 +3x+1−y=0 x=(((√(1+(((y−1)/2))^2 ))+((y−1)/2)))^(1/3) −(((√(1+(((y−1)/2))^2 ))−((y−1)/2)))^(1/3) f^(−1) (x)=(((√(1+(((x−1)/2))^2 ))+((x−1)/2)))^(1/3) −(((√(1+(((x−1)/2))^2 ))−((x−1)/2)))^(1/3) 1=3x^2 (dx/dy)+3(dx/dy) ⇒(dx/dy)=(1/(3(x^2 +1))) 0=6x((dx/dy))^2 +3x^2 (d^2 x/dy^2 )+3(d^2 x/dy^2 ) (d^2 x/dy^2 )=−((2x)/(9(x^2 +1)^3 )) ⇒(f^(−1) (x))′′=−((2f^(−1) (x))/(9(f^(−1) (x)^2 +1)^3 )) with f^(−1) (x)=(((√(1+(((x−1)/2))^2 ))+((x−1)/2)))^(1/3) −(((√(1+(((x−1)/2))^2 ))−((x−1)/2)))^(1/3) f^(−1) (5)=(((√(1+(((5−1)/2))^2 ))+((5−1)/2)))^(1/3) −(((√(1+(((5−1)/2))^2 ))−((5−1)/2)))^(1/3) f^(−1) (5)=(((√5)+2))^(1/3) −(((√5)−2))^(1/3) =1 (f^(−1) (5))′′=−((2×1)/(9(1^2 +1)^3 ))=−(1/(36))
$${f}\left({x}\right)={y}={x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{1} \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{1}−{y}=\mathrm{0} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{1}+\left(\frac{{y}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }+\frac{{y}−\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{1}+\left(\frac{{y}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }−\frac{{y}−\mathrm{1}}{\mathrm{2}}} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{1}+\left(\frac{{x}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }+\frac{{x}−\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{1}+\left(\frac{{x}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }−\frac{{x}−\mathrm{1}}{\mathrm{2}}} \\ $$$$ \\ $$$$\mathrm{1}=\mathrm{3}{x}^{\mathrm{2}} \frac{{dx}}{{dy}}+\mathrm{3}\frac{{dx}}{{dy}}\:\Rightarrow\frac{{dx}}{{dy}}=\frac{\mathrm{1}}{\mathrm{3}\left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\mathrm{0}=\mathrm{6}{x}\left(\frac{{dx}}{{dy}}\right)^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{2}} \frac{{d}^{\mathrm{2}} {x}}{{dy}^{\mathrm{2}} }+\mathrm{3}\frac{{d}^{\mathrm{2}} {x}}{{dy}^{\mathrm{2}} } \\ $$$$\frac{{d}^{\mathrm{2}} {x}}{{dy}^{\mathrm{2}} }=−\frac{\mathrm{2}{x}}{\mathrm{9}\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\Rightarrow\left({f}^{−\mathrm{1}} \left({x}\right)\right)''=−\frac{\mathrm{2}{f}^{−\mathrm{1}} \left({x}\right)}{\mathrm{9}\left({f}^{−\mathrm{1}} \left({x}\right)^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${with}\:{f}^{−\mathrm{1}} \left({x}\right)=\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{1}+\left(\frac{{x}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }+\frac{{x}−\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{1}+\left(\frac{{x}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }−\frac{{x}−\mathrm{1}}{\mathrm{2}}} \\ $$$${f}^{−\mathrm{1}} \left(\mathrm{5}\right)=\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{1}+\left(\frac{\mathrm{5}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }+\frac{\mathrm{5}−\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{1}+\left(\frac{\mathrm{5}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }−\frac{\mathrm{5}−\mathrm{1}}{\mathrm{2}}} \\ $$$${f}^{−\mathrm{1}} \left(\mathrm{5}\right)=\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}+\mathrm{2}}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}−\mathrm{2}}=\mathrm{1} \\ $$$$\left({f}^{−\mathrm{1}} \left(\mathrm{5}\right)\right)''=−\frac{\mathrm{2}×\mathrm{1}}{\mathrm{9}\left(\mathrm{1}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }=−\frac{\mathrm{1}}{\mathrm{36}} \\ $$
Commented by SANOGO last updated on 12/Sep/21
merci bien le dur
$${merci}\:{bien}\:{le}\:{dur} \\ $$
Answered by mnjuly1970 last updated on 12/Sep/21
(f^(−1) )′(5 )=(( 1)/(3x^2 +3)) ∣_( x=1) =(1/6) f ′ (x)= 3x^( 2) +3 (f^(−1) )′(y )=(1/(f ′(x))) y′ (f^( −1) )′′(y )= ((−f ′′(x))/(f ′^( 2) (x))) (f^(−1) )(5)= =((−6)/((f ′)^( 3) (1)))=((−6)/(216)) = ((−1)/(36))
$$\:\:\:\left({f}^{−\mathrm{1}} \right)'\left(\mathrm{5}\:\right)=\frac{\:\mathrm{1}}{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}}\:\mid_{\:{x}=\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\:\:\:\:\:{f}\:'\:\left({x}\right)=\:\mathrm{3}{x}^{\:\mathrm{2}} +\mathrm{3} \\ $$$$\:\:\:\:\:\:\left({f}\:^{−\mathrm{1}} \right)'\left({y}\:\right)=\frac{\mathrm{1}}{{f}\:'\left({x}\right)} \\ $$$$\:\:\:\:\:\:{y}'\:\left({f}^{\:−\mathrm{1}} \right)''\left({y}\:\right)=\:\frac{−{f}\:''\left({x}\right)}{{f}\:'^{\:\mathrm{2}} \left({x}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\left({f}\:^{−\mathrm{1}} \right)\left(\mathrm{5}\right)=\:=\frac{−\mathrm{6}}{\left({f}\:'\right)^{\:\mathrm{3}} \left(\mathrm{1}\right)}=\frac{−\mathrm{6}}{\mathrm{216}}\:=\:\frac{−\mathrm{1}}{\mathrm{36}} \\ $$
Commented by SANOGO last updated on 13/Sep/21
merci beaucoup le dur
$${merci}\:{beaucoup}\:{le}\:{dur} \\ $$

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