Question Number 41985 by Penguin last updated on 16/Aug/18
$$\mathrm{Prove}\:{e}^{{x}} \geqslant{x}+\mathrm{1}\:\forall{x}\in\mathbb{R}\:\mathrm{in}\:\mathrm{as}\:\mathrm{many}\:\mathrm{ways} \\ $$$$\mathrm{as}\:\mathrm{you}\:\mathrm{can}\:\mathrm{show} \\ $$
Commented by maxmathsup by imad last updated on 16/Aug/18
![let f(x) =e^x −x−1 we have f^′ (x) =e^x −1 so f^′ (x) =0 ⇔ x=0 and f^′ (x)≥0 ⇔x≥0 f is increasing on [0,+∞[ decreasing on]−∞,0] x −∞ 0 +∞ f^′ (x) − + f(x) decr incr we have f(0) =0 lim_(x→−∞) f(x) =lim_(x→−∞) (e^x −x−1) =+∞ lim_(x→+∞) f(x) =lim_(x→+∞) (e^x −x−1) =lim_(x→+∞) e^x =+∞ ⇒ f(x)≥0 ∀ x ∈ R ⇒ e^x ≥ x+1 ∀x∈ R .](https://www.tinkutara.com/question/Q42014.png)
$${let}\:{f}\left({x}\right)\:={e}^{{x}} −{x}−\mathrm{1}\:{we}\:{have}\: \\ $$$${f}^{'} \left({x}\right)\:={e}^{{x}} −\mathrm{1}\:\:{so}\:\:{f}^{'} \left({x}\right)\:=\mathrm{0}\:\Leftrightarrow\:{x}=\mathrm{0}\:\:{and}\:{f}^{'} \left({x}\right)\geqslant\mathrm{0}\:\Leftrightarrow{x}\geqslant\mathrm{0} \\ $$$${f}\:{is}\:{increasing}\:{on}\:\left[\mathrm{0},+\infty\left[\:\:{decreasing}\:{on}\right]−\infty,\mathrm{0}\right] \\ $$$$\:\:{x}\:\:\:\:\:\:\:\:\:\:−\infty\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:+\infty \\ $$$${f}^{'} \left({x}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:+ \\ $$$${f}\left({x}\right)\:\:\:\:\:\:\:\:\:\:\:\:{decr}\:\:\:\:\:\:\:\:{incr}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{we}\:{have}\:{f}\left(\mathrm{0}\right)\:=\mathrm{0} \\ $$$${lim}_{{x}\rightarrow−\infty} {f}\left({x}\right)\:={lim}_{{x}\rightarrow−\infty} \left({e}^{{x}} −{x}−\mathrm{1}\right)\:=+\infty \\ $$$${lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)\:={lim}_{{x}\rightarrow+\infty} \left({e}^{{x}} \:−{x}−\mathrm{1}\right)\:={lim}_{{x}\rightarrow+\infty} \:{e}^{{x}} \:\:=+\infty\:\:\:\Rightarrow \\ $$$${f}\left({x}\right)\geqslant\mathrm{0}\:\:\forall\:{x}\:\in\:{R}\:\Rightarrow\:{e}^{{x}} \geqslant\:{x}+\mathrm{1}\:\:\forall{x}\in\:{R}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18
$${p}={e}^{{x}} \\ $$$${q}={x}+\mathrm{1} \\ $$$${p}>{q}\:\:{if}\:\frac{{dp}}{{dx}}>\frac{{dq}}{{dx}} \\ $$$${now}\:\frac{{dp}}{{dx}}={e}^{{x}} \:\:\:\:\:\frac{{dq}}{{dx}}=\mathrm{1} \\ $$$${so}\:{e}^{{x}} −\mathrm{1}>\mathrm{0}\:\:\:{that}\:{means}\:{e}^{{x}} >{x}+\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$\frac{{d}}{{dx}}\left\{\frac{{dp}}{{dx}}−\frac{{dq}}{{dx}}\right\}={e}^{{x}} \\ $$$${when}\:{x}>\mathrm{0}\:\:\:{e}^{{x}} >\mathrm{0}\:{so}\:{curve}\:{hold}\:{water}\:{i},{e}\:{concave} \\ $$$${when}\:{x}<\mathrm{0}\:\:{e}^{{x}} >\mathrm{0}\:\:\:{so}\:{curve}\:{hold}\:{water}\:{i},{e}\:{concave} \\ $$$${that}\:{means}\:{e}^{{x}} >{x}+\mathrm{1} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18
Commented by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18
$${from}\:{graph}\:{it}\:{is}\:{clear}\:{that}\:{e}^{{x}} −{x}−\mathrm{1}>\mathrm{0}\:{when}\:{x}>\mathrm{0} \\ $$$${and}\:{when}\:{x}<\mathrm{0} \\ $$