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` .
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