We have $f:RR->RR,f(x)=e^(x-1)$ .How to demonstrate that $f(x)>x,$ for any $x inRR$ \{1}?

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Answer 1 · 2017-04-18T10:01:52

See below.

Calling $g(x) = f(x)-x = e^(x-1)-x$

we have

$g(1)=1-1=0$

and considering now $x = 1pm delta$ we have

$g(1pmdelta)=e^(pm delta)-(pmdelta)$

or

$g_1(1+delta)=e^delta-delta > 0$
$g_2(1-delta)=e^(-delta)+delta > 0$

so

$f(x) - x > 0$ for $x in RR,x ne 1$

We have f:RR->RR,f(x)=e^(x-1)f:RR->RR,f(x)=e^(x-1).How to demonstrate that f(x)>x,f(x)>x,for any x inRRx inRR\{1}?