Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 125815 by Tanuidesire last updated on 14/Dec/20

prove that the function f(x)=(1/(e^x −2))−2x, has a root in the open interval (0,1). recall that e≈2.7

$${prove}\:{that}\:{the}\:{function}\:{f}\left({x}\right)=\frac{\mathrm{1}}{{e}^{{x}} −\mathrm{2}}−\mathrm{2}{x},\:{has}\:{a}\:{root}\:{in}\:{the}\:{open}\:{interval}\:\left(\mathrm{0},\mathrm{1}\right).\:{recall}\:{that}\:{e}\approx\mathrm{2}.\mathrm{7} \\ $$

Answered by physicstutes last updated on 14/Dec/20

f (x) = (1/(e^x −2))−2x  f(0) = (1/(−1))−0 = −1  f(1) = (1/(e−2))−2 ≈−0.6  f(1)×f(0) > 0 hence f(x) does not have roots in that inteva

$${f}\:\left({x}\right)\:=\:\frac{\mathrm{1}}{{e}^{{x}} −\mathrm{2}}−\mathrm{2}{x} \\ $$$${f}\left(\mathrm{0}\right)\:=\:\frac{\mathrm{1}}{−\mathrm{1}}−\mathrm{0}\:=\:−\mathrm{1} \\ $$$${f}\left(\mathrm{1}\right)\:=\:\frac{\mathrm{1}}{{e}−\mathrm{2}}−\mathrm{2}\:\approx−\mathrm{0}.\mathrm{6} \\ $$$${f}\left(\mathrm{1}\right)×{f}\left(\mathrm{0}\right)\:>\:\mathrm{0}\:\mathrm{hence}\:{f}\left({x}\right)\:\mathrm{does}\:\mathrm{not}\:\mathrm{have}\:\mathrm{roots}\:\mathrm{in}\:\mathrm{that}\:\mathrm{inteva} \\ $$

Commented by Tanuidesire last updated on 14/Dec/20

thanks!

$${thanks}! \\ $$

Commented by mindispower last updated on 14/Dec/20

f(0).f(1)<0 withe contnuity ⇒f(x)=0 hase solution  f(0).f(1)>0 we can say nothing  exempl  f(x)=x+1,f(0).f(1)=2>0 f(x)=0 hase no solution  f(x)=(x−(1/2))^2   f(0)f(1)=(1/(16))>0 f(x)=0 hase solution x=(1/2)

$${f}\left(\mathrm{0}\right).{f}\left(\mathrm{1}\right)<\mathrm{0}\:{withe}\:{contnuity}\:\Rightarrow{f}\left({x}\right)=\mathrm{0}\:{hase}\:{solution} \\ $$$${f}\left(\mathrm{0}\right).{f}\left(\mathrm{1}\right)>\mathrm{0}\:{we}\:{can}\:{say}\:{nothing} \\ $$$${exempl} \\ $$$${f}\left({x}\right)={x}+\mathrm{1},{f}\left(\mathrm{0}\right).{f}\left(\mathrm{1}\right)=\mathrm{2}>\mathrm{0}\:{f}\left({x}\right)=\mathrm{0}\:{hase}\:{no}\:{solution} \\ $$$${f}\left({x}\right)=\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${f}\left(\mathrm{0}\right){f}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{16}}>\mathrm{0}\:{f}\left({x}\right)=\mathrm{0}\:{hase}\:{solution}\:{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com