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Question Number 124007 by liberty last updated on 30/Nov/20
Given a function y=f(x) where f^(−1) (((x+5)/(x−5)))=(8/(x+5)) Find slope of the curve y=f(x) at x=1 .
$$\:{Given}\:{a}\:{function}\:{y}={f}\left({x}\right)\:{where}\:{f}^{−\mathrm{1}} \left(\frac{{x}+\mathrm{5}}{{x}−\mathrm{5}}\right)=\frac{\mathrm{8}}{{x}+\mathrm{5}} \\ $$$${Find}\:{slope}\:{of}\:{the}\:{curve}\:{y}={f}\left({x}\right)\:{at}\:{x}=\mathrm{1}\:. \\ $$
Answered by john_santu last updated on 30/Nov/20
f^(−1) (((x+5)/(x−5)))=(8/(x+5)) ⇔ f((8/(x+5)))=((x+5)/(x−5)) differentiating both side give −(8/((x+5)^2 )) f ′((8/(x+5))) = −((10)/((x−5)^2 )) put x=3 ⇒ −(8/(64)) f ′(1) = −((10)/4) ⇒ f ′(1) = ((10)/4)×(8/1) = 20.
$${f}^{−\mathrm{1}} \left(\frac{{x}+\mathrm{5}}{{x}−\mathrm{5}}\right)=\frac{\mathrm{8}}{{x}+\mathrm{5}}\:\Leftrightarrow\:{f}\left(\frac{\mathrm{8}}{{x}+\mathrm{5}}\right)=\frac{{x}+\mathrm{5}}{{x}−\mathrm{5}} \\ $$$${differentiating}\:{both}\:{side}\:{give} \\ $$$$−\frac{\mathrm{8}}{\left({x}+\mathrm{5}\right)^{\mathrm{2}} }\:{f}\:'\left(\frac{\mathrm{8}}{{x}+\mathrm{5}}\right)\:=\:−\frac{\mathrm{10}}{\left({x}−\mathrm{5}\right)^{\mathrm{2}} } \\ $$$${put}\:{x}=\mathrm{3}\:\Rightarrow\:−\frac{\mathrm{8}}{\mathrm{64}}\:{f}\:'\left(\mathrm{1}\right)\:=\:−\frac{\mathrm{10}}{\mathrm{4}} \\ $$$$\Rightarrow\:{f}\:'\left(\mathrm{1}\right)\:=\:\frac{\mathrm{10}}{\mathrm{4}}×\frac{\mathrm{8}}{\mathrm{1}}\:=\:\mathrm{20}.\: \\ $$
Answered by mathmax by abdo last updated on 30/Nov/20
f^(−1) (((x+5)/(x−5)))=(8/(x+5)) ⇒f((8/(x+5)))=((x+5)/(x−5)) let (8/(x+5))=t ⇒tx+5t=8 ⇒ tx=8−5t ⇒x=((8−5t)/t) ⇒f(t)=((((8−5t)/t)+5)/(((8−5t)/t)−5))=((8t)/(8−10t)) =((4t)/(4−5t)) we have f(t)=((4t)/(4−5t)) ⇒f^′ (t)=((4(4−5t)−4t(−5))/((4−5t)^2 ))=((16−20t+20t)/((5t−4)^2 )) ⇒f^′ (t)=((16)/((5t−4)^2 )) ⇒f^′ (1) =16
$$\mathrm{f}^{−\mathrm{1}} \left(\frac{\mathrm{x}+\mathrm{5}}{\mathrm{x}−\mathrm{5}}\right)=\frac{\mathrm{8}}{\mathrm{x}+\mathrm{5}}\:\Rightarrow\mathrm{f}\left(\frac{\mathrm{8}}{\mathrm{x}+\mathrm{5}}\right)=\frac{\mathrm{x}+\mathrm{5}}{\mathrm{x}−\mathrm{5}}\:\:\mathrm{let}\:\frac{\mathrm{8}}{\mathrm{x}+\mathrm{5}}=\mathrm{t}\:\Rightarrow\mathrm{tx}+\mathrm{5t}=\mathrm{8}\:\Rightarrow \\ $$$$\mathrm{tx}=\mathrm{8}−\mathrm{5t}\:\Rightarrow\mathrm{x}=\frac{\mathrm{8}−\mathrm{5t}}{\mathrm{t}}\:\Rightarrow\mathrm{f}\left(\mathrm{t}\right)=\frac{\frac{\mathrm{8}−\mathrm{5t}}{\mathrm{t}}+\mathrm{5}}{\frac{\mathrm{8}−\mathrm{5t}}{\mathrm{t}}−\mathrm{5}}=\frac{\mathrm{8t}}{\mathrm{8}−\mathrm{10t}}\:=\frac{\mathrm{4t}}{\mathrm{4}−\mathrm{5t}} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{t}\right)=\frac{\mathrm{4t}}{\mathrm{4}−\mathrm{5t}}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{t}\right)=\frac{\mathrm{4}\left(\mathrm{4}−\mathrm{5t}\right)−\mathrm{4t}\left(−\mathrm{5}\right)}{\left(\mathrm{4}−\mathrm{5t}\right)^{\mathrm{2}} }=\frac{\mathrm{16}−\mathrm{20t}+\mathrm{20t}}{\left(\mathrm{5t}−\mathrm{4}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{f}^{'} \left(\mathrm{t}\right)=\frac{\mathrm{16}}{\left(\mathrm{5t}−\mathrm{4}\right)^{\mathrm{2}} }\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{1}\right)\:=\mathrm{16} \\ $$
Commented by liberty last updated on 30/Nov/20
not correct. f(t)=((((8−5t)/t)+5)/(((8−5t)/t)−5)) = ((8−5t+5t)/(8−5t−5t))=(8/(8−10t))=(4/(4−5t)) sir=
$${not}\:{correct}. \\ $$$${f}\left({t}\right)=\frac{\frac{\mathrm{8}−\mathrm{5}{t}}{{t}}+\mathrm{5}}{\frac{\mathrm{8}−\mathrm{5}{t}}{{t}}−\mathrm{5}}\:=\:\frac{\mathrm{8}−\mathrm{5}{t}+\mathrm{5}{t}}{\mathrm{8}−\mathrm{5}{t}−\mathrm{5}{t}}=\frac{\mathrm{8}}{\mathrm{8}−\mathrm{10}{t}}=\frac{\mathrm{4}}{\mathrm{4}−\mathrm{5}{t}} \\ $$$${sir}= \\ $$
Commented by Bird last updated on 30/Nov/20
sorry f(t)=(8/(8−10t)) =(4/(4−5t)) ⇒ f^′ (t)=4×(5/((4−5t)^2 ))=((20)/((4−5t)^2 )) ⇒ slope =f^′ (1)=20
$${sorry}\:{f}\left({t}\right)=\frac{\mathrm{8}}{\mathrm{8}−\mathrm{10}{t}}\:=\frac{\mathrm{4}}{\mathrm{4}−\mathrm{5}{t}}\:\Rightarrow \\ $$$${f}^{'} \left({t}\right)=\mathrm{4}×\frac{\mathrm{5}}{\left(\mathrm{4}−\mathrm{5}{t}\right)^{\mathrm{2}} }=\frac{\mathrm{20}}{\left(\mathrm{4}−\mathrm{5}{t}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${slope}\:={f}^{'} \left(\mathrm{1}\right)=\mathrm{20} \\ $$

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