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Question-206616




Question Number 206616 by universe last updated on 20/Apr/24
Answered by aleks041103 last updated on 21/Apr/24
e^(3x)  is increasing and continuous for x>0  ln(x) is incr. and cont. for x>0  ⇒f(x) is incr. and cont. for x>0  lim_(x→0)  f(x)→−∞  lim_(x→+∞)  f(x)→+∞  ⇒∃f^( −1) :R→R  f(f^( −1) (x))=x  ⇒(((df(y))/dy))_(y=f^( −1) (x)) ((df^( −1) (x))/dx) = 1 = (d/dx)x  ⇒((d f^( −1) (x))/dx) = (1/(f ′(f^( −1) (x))))  f^′ (x)=3e^(3x) +(1/x)  ⇒Df^( −1) (e^3 ) = (3 e^(3f^( −1) (e^3 )) +(1/(f^( −1) (e^3 ))))^(−1)   y=f^( −1) (e^3 )⇒f(y)=e^(3y) +ln(y)=e^3   ⇒y=1 is a soln.  since ∃!y:f(y)=e^3  ⇒ y=1=f^( −1) (e^3 )  ⇒Df^( −1) (e^3 ) = (1/(3e^(3.1) +(1/1)))=(1/(1+3e^3 ))
$${e}^{\mathrm{3}{x}} \:{is}\:{increasing}\:{and}\:{continuous}\:{for}\:{x}>\mathrm{0} \\ $$$${ln}\left({x}\right)\:{is}\:{incr}.\:{and}\:{cont}.\:{for}\:{x}>\mathrm{0} \\ $$$$\Rightarrow{f}\left({x}\right)\:{is}\:{incr}.\:{and}\:{cont}.\:{for}\:{x}>\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:{f}\left({x}\right)\rightarrow−\infty \\ $$$$\underset{{x}\rightarrow+\infty} {{lim}}\:{f}\left({x}\right)\rightarrow+\infty \\ $$$$\Rightarrow\exists{f}^{\:−\mathrm{1}} :\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({f}^{\:−\mathrm{1}} \left({x}\right)\right)={x} \\ $$$$\Rightarrow\left(\frac{{df}\left({y}\right)}{{dy}}\right)_{{y}={f}^{\:−\mathrm{1}} \left({x}\right)} \frac{{df}^{\:−\mathrm{1}} \left({x}\right)}{{dx}}\:=\:\mathrm{1}\:=\:\frac{{d}}{{dx}}{x} \\ $$$$\Rightarrow\frac{{d}\:{f}^{\:−\mathrm{1}} \left({x}\right)}{{dx}}\:=\:\frac{\mathrm{1}}{{f}\:'\left({f}^{\:−\mathrm{1}} \left({x}\right)\right)} \\ $$$${f}\:^{'} \left({x}\right)=\mathrm{3}{e}^{\mathrm{3}{x}} +\frac{\mathrm{1}}{{x}} \\ $$$$\Rightarrow{Df}^{\:−\mathrm{1}} \left({e}^{\mathrm{3}} \right)\:=\:\left(\mathrm{3}\:{e}^{\mathrm{3}{f}^{\:−\mathrm{1}} \left({e}^{\mathrm{3}} \right)} +\frac{\mathrm{1}}{{f}^{\:−\mathrm{1}} \left({e}^{\mathrm{3}} \right)}\right)^{−\mathrm{1}} \\ $$$${y}={f}^{\:−\mathrm{1}} \left({e}^{\mathrm{3}} \right)\Rightarrow{f}\left({y}\right)={e}^{\mathrm{3}{y}} +{ln}\left({y}\right)={e}^{\mathrm{3}} \\ $$$$\Rightarrow{y}=\mathrm{1}\:{is}\:{a}\:{soln}. \\ $$$${since}\:\exists!{y}:{f}\left({y}\right)={e}^{\mathrm{3}} \:\Rightarrow\:{y}=\mathrm{1}={f}^{\:−\mathrm{1}} \left({e}^{\mathrm{3}} \right) \\ $$$$\Rightarrow{Df}^{\:−\mathrm{1}} \left({e}^{\mathrm{3}} \right)\:=\:\frac{\mathrm{1}}{\mathrm{3}{e}^{\mathrm{3}.\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{1}}}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{3}{e}^{\mathrm{3}} } \\ $$

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