consider the function f(x) =`1(1+(1)/(x))^(x)`
The domain of f(x) is

Options

A. `(-1,0)cup(0,oo)`
B. R-{0}
C. `(-oo,-1)cup(0,oo)cup(0,oo)`
D. (0,oo)`

Answer

3

Explanation

`f(x)=1+(1)/(X)^(X)`
f(x) is defined if `1+1/xgt0 or (x+1)/(x)gt0`
or `(-oo,-1)cup(0,oo)`
Now f(X)=`(1+(1)/(x))^(x)ln (1+(1)/(x))+(x)/(1+(1)/(x))(-1)/(x^(2))`
`=(1+(1)/(x))^(x)ln(1+(1)/(x))-(1)/(x+1)`
Now `(1+(1)/(x))^(x)` is always positive Hence the sign of f(X)
Depends on the sign of ln `(1+(1)/(x))-(1)/(1+x)`
Let g(x)=`ln (1+(1)/(x))-(1)/(x+1)`
`g(x)=(1)/(1+(1)/(x))(-1)/(x^(2))+(1)/(x+1)^(2)=(-1)/(x(x+1)^(2)`
(i) for `x in (0,oo)g(x)lt0`
Thus g(X) is monotonically decreasing for x in `(0,oo)`
or `g(X) gt underset(xrarroo)limg(x)`
or `g(X)gt0, so f(X) gt0`
(ii) for `x in (-oo,-1),g(x)gt0`
Thus g(X) is monotonically increasing for `x in (-oo,-1)`
or `g(X) gt underset(xrarroo)limg(X)gt0`
`therefore f(x)gt0`
Hence from (i) nd (ii) we get
`f(X) gt0 forall x in (-oo,-1)cup(0,oo)`
Thus f(x) is montonically increasing in its domain Also
`underset(xrarroo)lim(1+(1)/(x))^(x)`=e
`underset(xrarr0)lim(1+(1)/(x))^(x)=1 and underset(xrarr-1)lim(1+(1)/(x))^(x)=oo`
The graph of f(X) is shown in figure

Range is `y in (1,oo)-{e}`