Draw the graph of the function `f(x) = (1+1/x)^(2)`

Explanation

`f(x) = (1+1/x)^(x),f(x)` is defined if `1+1/x gt 0`
`rArr (x+1)/x gt 0 rArr (-infty, -1) cup (0 , infty)`
Now `f^(')(x) =(1+1/x)^(x) ["ln "(1+1/x)+x/(1+1/x)-1/x^(2)]`
`=(1+1/x)(x) ["ln "(1+1/1x)-1/(x+1)]`
Now `(1+1/x)^(x)` is always positive, hence the sign of the `f^(')(x)` depends on the sign of `"ln "(1+1/x)-1/(x+1)`
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))`.
For `x in (0, infty), g^(')(x) lt 0`.
So `g(x)` is monotonically decreasing for `x in (0,infty)`. Thus,
`g(x) gt underset(x to infty)"lim"g(x)`
`rArr g(x) gt 0`
`rArr f^(')(x) gt 0`
For `x in (-infty, -1), g^(')(x) gt 0`.
So `g(x)` is monotonically increasing in its domain.
Also `underset( x to +-infty)"lim"(1+1/x)^(x) =e`
`underset(x to +-infty(1+1/x)^(x)=e`
`underset(x to 0^(+))"lim"(1+1/x)^(x)=1` and `underset(x to 1) "lim"(x+1/x)^(x)=infty`
From the above discussion the graph of the function is as shown in the following figure.

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