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).
Understand the Problem
The question is asking to analyze the function f(x) = x^2 + 1 in terms of its injectiveness and surjectiveness. We need to determine if the function is one-to-one (injective), onto (surjective), or both, and to provide a full explanation or proof for our conclusions.
Answer
Neither injective nor surjective.
Answer for screen readers
The function $f(x) = x^2 + 1$ is neither injective nor surjective.
Steps to Solve
- 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.
- 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.
- 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.
The function $f(x) = x^2 + 1$ is neither injective nor surjective.
More Information
The function $f(x) = x^2 + 1$ is a quadratic function that opens upwards. Its minimum value is 1, which occurs at $x = 0$. This means that while the function is not one-to-one or onto the entire set of real numbers, it does represent a parabolic shape with its vertex at (0,1).
Tips
- Confusing the roots of the equation with unique outputs. Remember that different inputs can yield the same output for non-injective functions.
- Assuming the function is surjective if it covers part of the codomain without confirming all possible outputs.
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