To find the domain of the function $f(x) = -x^2 + 1$, we need to determine the set of all possible input values (values of $x$) for which the function is defined.

This is a quadratic function, and since it's a polynomial function (specifically, a degree-2 polynomial), it is defined for all real numbers. There are no restrictions like division by zero or taking square roots of negative numbers.

Domain:

The domain of $f(x) = -x^2 + 1$ is all real numbers.

In interval notation, the domain is: $(-\infty, \infty)$

This can be verified graphically, where the graph of $f(x) = -x^2 + 1$ is a parabola opening downwards, and it extends indefinitely along the x-axis.

Would you like more details or have any questions?

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Here are 5 questions related to this concept:

1. What is the range of the function $f(x) = -x^2 + 1$?
2. How do we find the vertex of the quadratic function $f(x) = -x^2 + 1$?
3. How does the negative sign in front of $x^2$ affect the graph of the function?
4. Can the function $f(x) = -x^2 + 1$ ever have a positive slope?
5. What are the intercepts of the function $f(x) = -x^2 + 1$?

Tip: The domain of any polynomial function is always all real numbers unless there's an additional restriction or operation involved.