To verify if $f(x) = x^2 + 1$ and $g(x) = \sqrt{x^2 - 1}$ are inverse functions, we must check if $f(g(x)) = x$ and $g(f(x)) = x$.

Let's go through each calculation step-by-step.

Step 1: Calculate $f(g(x))$

1. Substitute $g(x) = \sqrt{x^2 - 1}$ into $f(x)$: $f(g(x)) = f\left(\sqrt{x^2 - 1}\right)$
2. Substitute $\sqrt{x^2 - 1}$ into the function $f(x) = x^2 + 1$: $f(g(x)) = \left(\sqrt{x^2 - 1}\right)^2 + 1$
3. Simplify: $f(g(x)) = x^2 - 1 + 1 = x^2$

So, $f(g(x)) \neq x$; instead, it equals $x^2$. This result indicates that $f(g(x)) \neq x$, which is a requirement for inverse functions.

Step 2: Calculate $g(f(x))$

1. Substitute $f(x) = x^2 + 1$ into $g(x)$: $g(f(x)) = g\left(x^2 + 1\right)$
2. Substitute $x^2 + 1$ into the function $g(x) = \sqrt{x^2 - 1}$: $g(f(x)) = \sqrt{(x^2 + 1)^2 - 1}$
3. Expand and simplify: g(f(x)) = \sqrt{x^4 + 2x^2 + 1 - 1} = \sqrt{x^4 + 2x^2} = x^2 \text{ (assuming x \geq 0)}

Again, $g(f(x)) \neq x$; it equals $x^2$, which means $g(f(x)) \neq x$.

Conclusion

Since neither $f(g(x)) = x$ nor $g(f(x)) = x$, the functions $f(x) = x^2 + 1$ and $g(x) = \sqrt{x^2 - 1}$ are not inverse functions.

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Do you have any questions or need further clarification?

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Further Questions

1. What are the properties of an inverse function?
2. How would you test if other functions are inverses of each other?
3. Can a function have more than one inverse?
4. Why is it necessary for both compositions $f(g(x)) = x$ and $g(f(x)) = x$ to be true?
5. What happens graphically when two functions are inverses?

Tip

When testing for inverse functions, always remember that both $f(g(x)) = x$ and $g(f(x)) = x$ must hold true for all $x$ in the domain.