For the function f(x) = x^2 + 1 from R to R, determine if it is injective, surjective, or both, and provide a full explanation (proof).

Steps to Solve

1. Check Injectiveness (One-to-One)

To determine if the function $f(x) = x^2 + 1$ is injective, we need to check if it assigns different outputs to different inputs. This can be done by setting the outputs equal:

$$ f(a) = f(b) $$

This leads to

$$ a^2 + 1 = b^2 + 1 $$

Then, simplify:

$$ a^2 = b^2 $$

Taking the square root gives us:

$$ a = b \text{ or } a = -b $$

Since $a = b$ does not necessarily mean the function is injective (because $f(a)$ could equal $f(-a)$ for $a \neq 0$), we conclude that $f(x)$ is not injective.

2. Check Surjectiveness (Onto)

Next, we need to check if the function covers all possible values in the codomain (output set). The range of $f(x) = x^2 + 1$ is everything greater than or equal to 1 since $x^2 \geq 0$ for all real $x$:

$$ f(x) \geq 1$$

Thus, the function does not cover all possible real numbers, only those in $[1, \infty)$. Therefore, $f(x)$ is not surjective either.

3. Conclusion on Injectiveness and Surjectiveness

Putting the results together, we can conclude that the function $f(x) = x^2 + 1$ is neither injective nor surjective.