Step 1: Check for one-one (injective)

A function \(f(x)\) is one-one if for any \(x_1, x_2\) in the domain, \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\).

Let \(f(x_1) = f(x_2)\). Then \(x_1^2 + 1 = x_2^2 + 1\).

This simplifies to \(x_1^2 = x_2^2\). Taking the square root, we get \(x_1 = \pm x_2\).

Since the domain is \(R_{+}\) (non-negative real numbers), \(x_1\) and \(x_2\) are non-negative. Therefore, \(x_1 = x_2\).

Thus, the function is one-one.

Step 2: Check for onto (surjective)

A function \(f(x)\) is onto if for every \(y\) in the codomain, there exists an \(x\) in the domain such that \(f(x) = y\).

Let \(y \in R_{+}\) be an element in the codomain. We want to find \(x \in R_{+}\) such that \(f(x) = y\).

So, \(x^2 + 1 = y\), which means \(x^2 = y - 1\), and \(x = \sqrt{y - 1}\).

Since \(x \in R_{+}\), we must have \(y - 1 \geq 0\), which means \(y \geq 1\).

However, the codomain is \(R_{+}\), which includes values between 0 and 1. For example, if \(y = 0.5\), then \(x = \sqrt{0.5 - 1} = \sqrt{-0.5}\), which is not a real number.

Therefore, the function is not onto.