You need to remember that you may find the equation of the tangent line to the graph of `f(x),` at a point `x = x_0` , using the following formula such that:

`y - f(x_0) = f'(x_0)(x - x_0)`

Selecting a value in the given interval `[0.5,2],` `x_0 = 1` , the equation...

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You need to remember that you may find the equation of the tangent line to the graph of `f(x),` at a point `x = x_0` , using the following formula such that:

`y - f(x_0) = f'(x_0)(x - x_0)`

Selecting a value in the given interval `[0.5,2],` `x_0 = 1` , the equation of the tangent line is:

`y - f(1) = f'(1)(x - 1)`

You need to find `f'(x)` such that:

`f'(x) = (x + 1/x)' => f'(x) = 1 - 1/x^2`

Substituting 1 for x yields:

`f'(1) = 1 - 1/1 = 1-1 = 0`

You need to evaluate `f(1) ` such that:

`f(1) = 1 + 1/1 = 2`

Substituting 2 for `f(1)` and 0 for `f'(1)` yields:

`y - 2 = 0*(x - 1) => y = 2`

**Hence, evaluating the equation of the tangent line to the graph of the given function, at a point in the given interval `[0.5,2]` yields `y = 2` .**