You need to prove that the function is both injective and surjective.

You should remember that a function is injective if for `x_1!=x_2 =gt f(x_1)!=f(x_2). `

You may notice the monotony of a function using its derivative such that:

`f'(x) = 2x`

Notice that substituting values from domain of function `(1,+oo)` for...

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You need to prove that the function is both injective and surjective.

You should remember that a function is injective if for `x_1!=x_2 =gt f(x_1)!=f(x_2). `

You may notice the monotony of a function using its derivative such that:

`f'(x) = 2x`

Notice that substituting values from domain of function `(1,+oo)` for x yields that the function increases over `(2,+oo), ` hence if `x_1ltx_2` => `f(x_1)ltf(x_2)` = > the function is injective.

You need to verify if the function is surjective, hence you need to solve for x the equation `y = x^2 + 1` such that:

`x^2 = y - 1=gt x = sqrt(y-1)` (you need to consider only positive value of square root since `x in ( 1, +oo))`

Since `y in (2,+oo)=gt y-1 gt1 gt 0` => the square root is defined, hence for each value of `y in (2,+oo)` , there is at least one correspondent value of `x in (1,+oo).`

Hence, the function f(x) is surjective.

**Hence, since the function is both injective and surjective over the given domain and range, then the function is a bijection.**