Question Number 123752 by liberty last updated on 27/Nov/20
$${Let}\:{f}\left({x}\right)\:{be}\:{differentiable}\:{function} \\ $$$${and}\:{g}\left({x}\right)\:{be}\:{the}\:{inverse}\:{of}\:{f}\left({x}\right). \\ $$$${when}\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\left(\frac{{f}\left({x}\right)−{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} −\mathrm{4}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:{then}\:\frac{{dg}\left({x}\right)}{{dx}}\mid_{{x}=\mathrm{8}} \:=? \\ $$
Commented by benjo_mathlover last updated on 28/Nov/20
Answered by ajfour last updated on 27/Nov/20
$${say}\:\:{f}\left({x}\right)−{x}^{\mathrm{3}} =\mathrm{2}\left({x}−\mathrm{2}\right) \\ $$$$\Rightarrow\:\:\:{f}\left({x}\right)={x}^{\mathrm{3}} +\mathrm{2}{x}−\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:{x}={g}^{\mathrm{3}} +\mathrm{2}{g}−\mathrm{4} \\ $$$${for}\:\:{x}=\mathrm{8}\:\:\:\:\:{g}^{\mathrm{3}} +\mathrm{2}{g}−\mathrm{12}=\mathrm{0} \\ $$$$\:\:\:\:\Rightarrow\:{g}=\mathrm{2} \\ $$$$\:\:\:\:\mathrm{1}=\mathrm{3}{g}^{\mathrm{2}} \left(\frac{{dg}}{{dX}}\right)+\mathrm{2}\left(\frac{{dg}}{{dx}}\right) \\ $$$$\Rightarrow\:\:\left(\frac{{dg}}{{dx}}\right)\mid_{{x}=\mathrm{8}} \:=\:\frac{\mathrm{1}}{\mathrm{3}{g}^{\mathrm{2}} +\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{14}}\:. \\ $$$$ \\ $$
Answered by bobhans last updated on 27/Nov/20
$${g}\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right)\:{then}\:{g}\:'\left({x}\right)\:=\:\left({f}^{−\mathrm{1}} \right)'\left({x}\right) \\ $$$${from}\:{the}\:{limit}\:{we}\:{get}\:{f}\left(\mathrm{2}\right)=\mathrm{8}\: \\ $$$${or}\:{f}^{−\mathrm{1}} \left(\mathrm{8}\right)\:=\:\mathrm{2}\: \\ $$$${we}\:{know}\:{that}\:{g}'\left(\mathrm{8}\right)=\frac{\mathrm{1}}{{f}\:'\left({f}^{−\mathrm{1}} \left(\mathrm{8}\right)\right)}...\left(\bullet\right) \\ $$$$ \\ $$$${consider}\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{{f}\left({x}\right)−{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} −\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{f}\:'\left({x}\right)−\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}{x}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Leftrightarrow\:{f}\:'\left(\mathrm{2}\right)−\mathrm{12}=\mathrm{2}\:;\:{f}\:'\left(\mathrm{2}\right)=\mathrm{14} \\ $$$${substitute}\:{into}\:\left(\bullet\right)\:{we}\:{get} \\ $$$${g}'\left(\mathrm{8}\right)=\frac{\mathrm{1}}{{f}\:'\left({f}^{−\mathrm{1}} \left(\mathrm{8}\right)\right)}=\frac{\mathrm{1}}{{f}\:'\left(\mathrm{2}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{14}}. \\ $$$$ \\ $$