A function is one-one if for every \(x_1, x_2\) in the domain, \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\). Let \(f(x_1) = f(x_2)\). Then, \[x_1^2 + 1 = x_2^2 + 1\] \[x_1^2 = x_2^2\] Since the domain is \(R_{+}\) (non-negative real numbers), taking the square root gives \[x_1 = x_2\] Thus, the function is one-one.
A function 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 \(f(x) = y\). \[x^2 + 1 = y\] \[x^2 = y - 1\] \[x = \sqrt{y - 1}\] Since \(x\) must be in \(R_{+}\), we need \(y - 1 \geq 0\), 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.
The function is one-one but not onto.
Final Answer: one-one but not onto
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