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Question Number 56356 by Tawa1 last updated on 15/Mar/19
Find the minimum value of the function F(x) = log_e x − x for x > 0. Hence show that: log_e x ≤ x − 1. For all x > 0
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:\mathrm{F}\left(\mathrm{x}\right)\:=\:\:\mathrm{log}_{\mathrm{e}} \mathrm{x}\:\:−\:\:\mathrm{x}\:\:\:\mathrm{for}\:\:\:\mathrm{x}\:>\:\mathrm{0}. \\ $$$$\:\:\mathrm{Hence}\:\mathrm{show}\:\mathrm{that}:\:\:\:\mathrm{log}_{\mathrm{e}} \mathrm{x}\:\:\leqslant\:\:\mathrm{x}\:−\:\mathrm{1}.\:\:\mathrm{For}\:\mathrm{all}\:\:\:\mathrm{x}\:>\:\mathrm{0} \\ $$
Commented by Tawa1 last updated on 15/Mar/19
Answered by kaivan.ahmadi last updated on 15/Mar/19
f(x)=lnx−x⇒f′(x)=(1/x)−1=0⇒x=1 if x<1⇒(1/x)>1⇒(1/x)−1>0⇒f′(x)>0 somillarly if x>1⇒f′(x)<0 ⇒f(1)=−1 is maximum⇒ f(x)≤−1⇒lnx−x≤−1⇒ lnx≤x−1
$${f}\left({x}\right)={lnx}−{x}\Rightarrow{f}'\left({x}\right)=\frac{\mathrm{1}}{{x}}−\mathrm{1}=\mathrm{0}\Rightarrow{x}=\mathrm{1} \\ $$$${if}\:{x}<\mathrm{1}\Rightarrow\frac{\mathrm{1}}{{x}}>\mathrm{1}\Rightarrow\frac{\mathrm{1}}{{x}}−\mathrm{1}>\mathrm{0}\Rightarrow{f}'\left({x}\right)>\mathrm{0} \\ $$$${somillarly}\:{if}\:{x}>\mathrm{1}\Rightarrow{f}'\left({x}\right)<\mathrm{0} \\ $$$$\Rightarrow{f}\left(\mathrm{1}\right)=−\mathrm{1}\:{is}\:{maximum}\Rightarrow \\ $$$${f}\left({x}\right)\leqslant−\mathrm{1}\Rightarrow{lnx}−{x}\leqslant−\mathrm{1}\Rightarrow \\ $$$${lnx}\leqslant{x}−\mathrm{1} \\ $$$$ \\ $$$$ \\ $$
Commented by Tawa1 last updated on 15/Mar/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Mar/19
((df(x))/dx)=(1/x)−1 (d^2 f/dx^2 )=((−1)/x^2 ) for min/max (df/dx)=0 so (1/x)−1=0 x=1 at x=1 (d^2 f/dx^2 )=((−1)/1^2 )<0 so no minimum value... for min (d^2 f/dx^2 )>0 and for max (d^2 f/dx^2 )<0 so at x=1 f(x)=∣lnx−x∣_(x=1) →−1 value=−1
$$\frac{{df}\left({x}\right)}{{dx}}=\frac{\mathrm{1}}{{x}}−\mathrm{1} \\ $$$$\frac{{d}^{\mathrm{2}} {f}}{{dx}^{\mathrm{2}} }=\frac{−\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$${for}\:{min}/{max}\:\:\:\frac{{df}}{{dx}}=\mathrm{0} \\ $$$${so}\:\frac{\mathrm{1}}{{x}}−\mathrm{1}=\mathrm{0} \\ $$$${x}=\mathrm{1} \\ $$$${at}\:{x}=\mathrm{1}\:\frac{{d}^{\mathrm{2}} {f}}{{dx}^{\mathrm{2}} }=\frac{−\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }<\mathrm{0} \\ $$$${so}\:\:{no}\:{minimum}\:\:{value}… \\ $$$${for}\:{min}\:\frac{{d}^{\mathrm{2}} {f}}{{dx}^{\mathrm{2}} }>\mathrm{0}\:\:\:{and}\:\:{for}\:{max}\:\frac{{d}^{\mathrm{2}} {f}}{{dx}^{\mathrm{2}} }<\mathrm{0} \\ $$$${so}\:{at}\:{x}=\mathrm{1}\:{f}\left({x}\right)=\mid{lnx}−{x}\mid_{{x}=\mathrm{1}} \rightarrow−\mathrm{1} \\ $$$$\:{value}=−\mathrm{1} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 15/Mar/19
Commented by tanmay.chaudhury50@gmail.com last updated on 15/Mar/19
from graph of lnx−x it is clear that f(x) hss max value at x=1 and that value is −1
$${from}\:{graph}\:{of}\:{lnx}−{x}\:{it}\:{is}\:{clear}\:{that}\:{f}\left({x}\right)\:{hss}\:{max}\:{value}\:{at}\:{x}=\mathrm{1}\:{and}\:{that}\:{value}\:{is}\:−\mathrm{1} \\ $$
Commented by Tawa1 last updated on 15/Mar/19
God bless you sir. i appreciate your effort
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Mar/19
if (dg/dx)<(dh/dx) then g(x)<h(x) now (d/dx)(lnx)=(1/x) (d/dx)(x−1)=1 now (1/x)−1 ((1−x)/x)<0 when x>1 so lnx<(x−1) when x>1 at x=1 i,e ln(1)=0 (1−1)=0 (lnx)_(x=1) =(x−1)_(x=1) when 1>x>0 lnx=−ve (x−1)=−ve and ∣lnx∣>∣(x−1)∣ so lnx<(x−1) when 1>x>0 so lnx<(x−1) when x>0
$${if}\:\frac{{dg}}{{dx}}<\frac{{dh}}{{dx}} \\ $$$${then}\:{g}\left({x}\right)<{h}\left({x}\right) \\ $$$${now}\:\frac{{d}}{{dx}}\left({lnx}\right)=\frac{\mathrm{1}}{{x}} \\ $$$$\frac{{d}}{{dx}}\left({x}−\mathrm{1}\right)=\mathrm{1} \\ $$$${now}\:\frac{\mathrm{1}}{{x}}−\mathrm{1} \\ $$$$\frac{\mathrm{1}−{x}}{{x}}<\mathrm{0}\:{when}\:{x}>\mathrm{1} \\ $$$${so}\:{lnx}<\left({x}−\mathrm{1}\right)\:\:{when}\:{x}>\mathrm{1} \\ $$$${at}\:{x}=\mathrm{1}\:\:{i},{e}\:{ln}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\left(\mathrm{1}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({lnx}\right)_{{x}=\mathrm{1}} =\left({x}−\mathrm{1}\right)_{{x}=\mathrm{1}} \\ $$$${when}\:\mathrm{1}>{x}>\mathrm{0}\:\: \\ $$$${lnx}=−{ve}\:\:\:\:\left({x}−\mathrm{1}\right)=−{ve} \\ $$$${and}\:\:\:\mid{lnx}\mid>\mid\left({x}−\mathrm{1}\right)\mid \\ $$$${so}\:{lnx}<\left({x}−\mathrm{1}\right)\:{when}\:\mathrm{1}>{x}>\mathrm{0} \\ $$$${so}\:{lnx}<\left({x}−\mathrm{1}\right)\:{when}\:{x}>\mathrm{0} \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 15/Mar/19
Commented by Tawa1 last updated on 15/Mar/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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