Let us see.

f'(x) = 1 -1/(x^2)

Set that to zero,

0 = 1 -1/(x^2)

Clear the fractions, multiply both sides by x^2

0 = x^2 -1

0 = (x+1)(x-1)

So,

x = 1 or (-1) -----------------**

Or,

0 = x^2 -1

x^2 = 1

x = +,-sqrt(1)

x = 1 or (-1) --------------same.

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f''(x) = 2/(x^3)

Set that to 0,

0 = 2/(x^3)

Clear the fractions, multiply both sides by x^3,

0 = 2 -------------yeah, right.

Meaning, there is no inflection point.

You have and know how to use a graphing calculator? See the graph of f(x) = 1 +1/x. Maybe the graph really does not have an inflection point.