Prove e^x ≥x+1 ∀x∈R in as many ways as you can show


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 41985 by Penguin last updated on 16/Aug/18

$Prove e^{x} ⩾ x + 1 \forall x \in R in as many ways$ $as you can show$
Commented by maxmathsup by imad last updated on 16/Aug/18

$l e t f (x) = e^{x} - x - 1 w e h a v e$ $f^{^{'}} (x) = e^{x} - 1 s o f^{^{'}} (x) = 0 \Leftrightarrow x = 0 a n d f^{^{'}} (x) ⩾ 0 \Leftrightarrow x ⩾ 0$ $f i s i n c r e a s i n g o n [0, + \infty [d e c r e a s i n g o n] - \infty, 0]$ $x - \infty 0 + \infty$ $f^{^{'}} (x) - +$ $f (x) d e c r i n c r w e h a v e f (0) = 0$ ${l i m}_{x \to - \infty} f (x) = {l i m}_{x \to - \infty} (e^{x} - x - 1) = + \infty$ ${l i m}_{x \to + \infty} f (x) = {l i m}_{x \to + \infty} (e^{x} - x - 1) = {l i m}_{x \to + \infty} e^{x} = + \infty \Rightarrow$ $f (x) ⩾ 0 \forall x \in R \Rightarrow e^{x} ⩾ x + 1 \forall x \in R .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18
	$p=e^x q=x+1 p>q if (dp/dx)>(dq/dx) now (dp/dx)=e^x (dq/dx)=1 so e^x −1>0 that means e^x >x+1 (d/dx){(dp/dx)−(dq/dx)}=e^x when x>0 e^x >0 so curve hold water i,e concave when x<0 e^x >0 so curve hold water i,e concave that means e^x >x+1$
	$p = e^{x}$ $q = x + 1$ $p > q i f \frac{d p}{d x} > \frac{d q}{d x}$ $n o w \frac{d p}{d x} = e^{x} \frac{d q}{d x} = 1$ $s o e^{x} - 1 > 0 t h a t m e a n s e^{x} > x + 1$ $\frac{d}{d x} {\frac{d p}{d x} - \frac{d q}{d x}} = e^{x}$ $w h e n x > 0 e^{x} > 0 s o c u r v e h o l d w a t e r i, e c o n c a v e$ $w h e n x < 0 e^{x} > 0 s o c u r v e h o l d w a t e r i, e c o n c a v e$ $t h a t m e a n s e^{x} > x + 1$
	Commented by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18

	$f r o m g r a p h i t i s c l e a r t h a t e^{x} - x - 1 > 0 w h e n x > 0$ $a n d w h e n x < 0$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum