# How to show x<e^x-1 if x<-1?

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You need to isolate `e^x` to one side, hence you should add 1 both sides such that:

`x + 1 lt e^x`

Dividing both sides by x + 1 yields:

`1 lt e^x/(x+1) \implies e^x/(x+1) - 1 gt 0 \implies (e^x - (x+1))/(x+1) gt 0`

You need to prove that the fraction `(e^x - (x+1))/(x+1)` is positive if `x > -1` .

You should differentiate this fraction to check if the derivative increases or decreases.

You need to use the quotient rule such that:

`[(e^x - (x+1))/(x+1)]' = (e^x(x+1) - e^x)/(x+1)^2`

`[(e^x - (x+1))/(x+1)]' = e^x(x + 1 - 1)/(x+1)^2 \implies [(e^x - (x+1))/(x+1)]' = xe^x/(x+1)^2`

`` **The denominator and the numerator of derivative are positive all over the interval (-1 ; oo), hence the derivative increases =gt the fraction is positive** `e^x/(x+1) - 1 gt 0` .