Menu Close

A-differentiable-function-f-x-satisfies-f-x-3-x-2-x-2-x-1-for-every-real-number-x-When-g-x-is-the-inverse-function-of-f-x-find-g-4-

Tinku Tara 0 Comments
Pin




Question Number 103322 by abony1303 last updated on 14/Jul/20
A differentiable function f(x) satisfies f(x^3 −x^2 +x)=2^(x+1) for every real number x. When g(x) is the inverse function of f(x), find g′(4)?
$$\mathrm{A}\:\mathrm{differentiable}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{satisfies} \\ $$$$\mathrm{f}\left(\mathrm{x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{2}} +\mathrm{x}\right)=\mathrm{2}^{\mathrm{x}+\mathrm{1}} \:\:\mathrm{for}\:\mathrm{every}\:\mathrm{real}\:\mathrm{number}\:{x}. \\ $$$$\mathrm{When}\:\mathrm{g}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{inverse}\:\mathrm{function}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right), \\ $$$$\mathrm{find}\:\mathrm{g}'\left(\mathrm{4}\right)? \\ $$
Commented by abony1303 last updated on 14/Jul/20
pls help
$$\mathrm{pls}\:\mathrm{help} \\ $$
Answered by Worm_Tail last updated on 14/Jul/20
f(x^3 −x^2 +x)=2^(x+1) ⇒ x^3 −x^2 +x=f^(−1) (2^(x+1) ) x^3 −x^2 +x=g(2^(x+1) ) 3x^2 −2x+1=2^(x+1) ln(2)g′(2^(x+1) )_(x=1) 3−2+1=4ln(2)g′(4) 2=4ln(2)g′(4)⇒g′(4)=(1/(2ln(2)))
$${f}\left({x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{x}\right)=\mathrm{2}^{{x}+\mathrm{1}} \Rightarrow \\ $$$${x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{x}={f}^{−\mathrm{1}} \left(\mathrm{2}^{{x}+\mathrm{1}} \right) \\ $$$${x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{x}={g}\left(\mathrm{2}^{{x}+\mathrm{1}} \right) \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}=\mathrm{2}^{{x}+\mathrm{1}} {ln}\left(\mathrm{2}\right){g}'\left(\mathrm{2}^{{x}+\mathrm{1}} \right)_{{x}=\mathrm{1}} \\ $$$$\mathrm{3}−\mathrm{2}+\mathrm{1}=\mathrm{4}{ln}\left(\mathrm{2}\right){g}'\left(\mathrm{4}\right) \\ $$$$\mathrm{2}=\mathrm{4}{ln}\left(\mathrm{2}\right){g}'\left(\mathrm{4}\right)\Rightarrow{g}'\left(\mathrm{4}\right)=\frac{\mathrm{1}}{\mathrm{2}{ln}\left(\mathrm{2}\right)} \\ $$
Commented by abony1303 last updated on 14/Jul/20
Thank you sir
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by mathmax by abdo last updated on 14/Jul/20
we have f(x^3 −x^2 +x) =2^(x+1) ⇒f^(−1) (2^(x+1) ) =x^3 −x^2 +x let 2^(x+1) =t ⇒e^((x+1)ln2) =t ⇒(x+1)ln2 =ln(t) ⇒x+1 =((lnt)/(ln2)) ⇒ x =((lnt)/(ln2))−1 ⇒f^(−1) (t) =(((lnt)/(ln2)))^3 −(((lnt)/(ln2)))^2 +((lnt)/(ln2)) =g(t) ⇒ g^′ (t) =(1/((ln2)^3 ))×3(((lnt)^2 )/t) −(1/((ln2)^2 ))×2((lnt)/t) +(1/(tln2)) ⇒ g^′ (4) =(3/(4(ln2)^3 ))(2ln2)^2 −(2/((ln2)^2 ))×((2ln(2))/4) +(1/(4ln2)) =(3/(ln2))−(1/(ln2)) +(1/(4ln2)) =(2/(ln2)) +(1/(4ln2)) =(2+(1/4))×(1/(ln2)) =(9/(4ln2))
$$\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{2}} \:+\mathrm{x}\right)\:=\mathrm{2}^{\mathrm{x}+\mathrm{1}} \:\Rightarrow\mathrm{f}^{−\mathrm{1}} \left(\mathrm{2}^{\mathrm{x}+\mathrm{1}} \right)\:=\mathrm{x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{2}} +\mathrm{x} \\ $$$$\mathrm{let}\:\mathrm{2}^{\mathrm{x}+\mathrm{1}} \:=\mathrm{t}\:\Rightarrow\mathrm{e}^{\left(\mathrm{x}+\mathrm{1}\right)\mathrm{ln2}} \:=\mathrm{t}\:\Rightarrow\left(\mathrm{x}+\mathrm{1}\right)\mathrm{ln2}\:=\mathrm{ln}\left(\mathrm{t}\right)\:\Rightarrow\mathrm{x}+\mathrm{1}\:=\frac{\mathrm{lnt}}{\mathrm{ln2}}\:\Rightarrow \\ $$$$\mathrm{x}\:=\frac{\mathrm{lnt}}{\mathrm{ln2}}−\mathrm{1}\:\Rightarrow\mathrm{f}^{−\mathrm{1}} \left(\mathrm{t}\right)\:=\left(\frac{\mathrm{lnt}}{\mathrm{ln2}}\right)^{\mathrm{3}} −\left(\frac{\mathrm{lnt}}{\mathrm{ln2}}\right)^{\mathrm{2}} \:+\frac{{lnt}}{{ln}\mathrm{2}}\:={g}\left({t}\right)\:\Rightarrow \\ $$$$\mathrm{g}^{'} \left(\mathrm{t}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{ln2}\right)^{\mathrm{3}} }×\mathrm{3}\frac{\left(\mathrm{lnt}\right)^{\mathrm{2}} }{\mathrm{t}}\:−\frac{\mathrm{1}}{\left(\mathrm{ln2}\right)^{\mathrm{2}} }×\mathrm{2}\frac{\mathrm{lnt}}{\mathrm{t}}\:+\frac{\mathrm{1}}{\mathrm{tln2}}\:\Rightarrow \\ $$$$\mathrm{g}^{'} \left(\mathrm{4}\right)\:=\frac{\mathrm{3}}{\mathrm{4}\left(\mathrm{ln2}\right)^{\mathrm{3}} }\left(\mathrm{2ln2}\right)^{\mathrm{2}} \:\:−\frac{\mathrm{2}}{\left(\mathrm{ln2}\right)^{\mathrm{2}} }×\frac{\mathrm{2ln}\left(\mathrm{2}\right)}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{4ln2}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{ln2}}−\frac{\mathrm{1}}{\mathrm{ln2}}\:+\frac{\mathrm{1}}{\mathrm{4ln2}}\:=\frac{\mathrm{2}}{\mathrm{ln2}}\:+\frac{\mathrm{1}}{\mathrm{4ln2}}\:=\left(\mathrm{2}+\frac{\mathrm{1}}{\mathrm{4}}\right)×\frac{\mathrm{1}}{\mathrm{ln2}}\:=\frac{\mathrm{9}}{\mathrm{4ln2}} \\ $$
Commented by 1549442205 last updated on 15/Jul/20
Sir′s the idea is right ,however mistaked at third line:f^(−1) (t)=(((lnt)/(ln2))−1)^3 −(((lnt)/(ln2))−1)^2 +((lnt)/(ln2))−1 =(((lnt)/(ln2)))^3 −3(((lnt)/(ln2)))^2 +3((lnt)/(ln2))−1−[(((lnt)/(ln2)))^2 −2((lnt)/(ln2))+1]+((lnt)/(ln2))−1 =(((lnt)/(ln2)))^3 −4(((lnt)/(ln2)))^2 +6((lnt)/(ln2))−2 g′(t)=(1/((ln2)^3 ))×3(((lnt)^2 )/t)−(4/((ln2)^2 ))×2((lnt)/t)+(6/(tln2)) g′(4)=(3/(ln2))−(4/(ln2))+(3/(2ln2))=(1/(2ln2))
$$\mathrm{Sir}'\mathrm{s}\:\mathrm{the}\:\mathrm{idea}\:\mathrm{is}\:\mathrm{right}\:,\mathrm{however} \\ $$$$\mathrm{mistaked}\:\:\mathrm{at}\:\mathrm{third}\:\mathrm{line}:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{t}\right)=\left(\frac{\mathrm{lnt}}{\mathrm{ln2}}−\mathrm{1}\right)^{\mathrm{3}} −\left(\frac{\mathrm{lnt}}{\mathrm{ln2}}−\mathrm{1}\right)^{\mathrm{2}} +\frac{\mathrm{lnt}}{\mathrm{ln2}}−\mathrm{1} \\ $$$$=\left(\frac{\mathrm{lnt}}{\mathrm{ln2}}\right)^{\mathrm{3}} −\mathrm{3}\left(\frac{\mathrm{lnt}}{\mathrm{ln2}}\right)^{\mathrm{2}} +\mathrm{3}\frac{\mathrm{lnt}}{\mathrm{ln2}}−\mathrm{1}−\left[\left(\frac{\mathrm{lnt}}{\mathrm{ln2}}\right)^{\mathrm{2}} −\mathrm{2}\frac{\mathrm{lnt}}{\mathrm{ln2}}+\mathrm{1}\right]+\frac{\mathrm{lnt}}{\mathrm{ln2}}−\mathrm{1} \\ $$$$=\left(\frac{\mathrm{lnt}}{\mathrm{ln2}}\right)^{\mathrm{3}} −\mathrm{4}\left(\frac{\mathrm{lnt}}{\mathrm{ln2}}\right)^{\mathrm{2}} +\mathrm{6}\frac{\mathrm{lnt}}{\mathrm{ln2}}−\mathrm{2} \\ $$$$\mathrm{g}'\left(\mathrm{t}\right)=\frac{\mathrm{1}}{\left(\mathrm{ln2}\right)^{\mathrm{3}} }×\mathrm{3}\frac{\left(\mathrm{lnt}\right)^{\mathrm{2}} }{\mathrm{t}}−\frac{\mathrm{4}}{\left(\mathrm{ln2}\right)^{\mathrm{2}} }×\mathrm{2}\frac{\mathrm{lnt}}{\mathrm{t}}+\frac{\mathrm{6}}{\mathrm{tln2}} \\ $$$$\mathrm{g}'\left(\mathrm{4}\right)=\frac{\mathrm{3}}{\mathrm{ln2}}−\frac{\mathrm{4}}{\mathrm{ln2}}+\frac{\mathrm{3}}{\mathrm{2ln2}}=\frac{\mathrm{1}}{\mathrm{2ln2}} \\ $$
Commented by mathmax by abdo last updated on 14/Jul/20
yes thank you sir...
$$\mathrm{yes}\:\:\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}… \\ $$
Commented by mathmax by abdo last updated on 14/Jul/20
i forget −1!
$$\mathrm{i}\:\mathrm{forget}\:−\mathrm{1}! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *