In general, f^(-1)(f(x))=f(f^(-1)(x))=

Property Explanation: The problem states the general property of a function and its inverse: f^(-1)(f(x))=f(f^(-1)(x)). This property tells us that when you apply a function to an input and then apply the inverse function to the result, you should get the original input back. Similarly, if you apply the inverse function to an input and then apply the function to the result, you should also get the original input back. This is because the function and its inverse undo each other's operations.Verification Process: To verify this property, let's consider what happens when we apply f to x. We get some output y, such that y = f(x). Now, if we apply the inverse function f^(-1) to y, we should get back our original input x, because the inverse function undoes the action of the function. So, f^(-1)(f(x)) = x.Conclusion: Similarly, if we start with some value x and apply the inverse function f^(-1), we get some output y such that y = f^(-1)(x). Now, if we apply the function f to y, we should get back our original input x, because the function undoes the action of the inverse function. So, f(f^(-1)(x)) = x.Since both f^(-1)(f(x)) and f(f^(-1)(x)) result in x, we can conclude that the property f^(-1)(f(x))=f(f^(-1)(x)) holds true for any function f and its inverse f^(-1), provided that f is invertible and the domain and codomain are appropriately defined so that the compositions are valid.

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In general, $f^{-1}(f(x))=f\left(f^{-1}(x)\right)=$

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Algebra 2

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Q. In general,

f^{-1}(f(x))=f\left(f^{-1}(x)\right)=

Property Explanation: The problem states the general property of a function and its inverse: $f^{-1}(f(x))=f(f^{-1}(x))$ . This property tells us that when you apply a function to an input and then apply the inverse function to the result, you should get the original input back. Similarly, if you apply the inverse function to an input and then apply the function to the result, you should also get the original input back. This is because the function and its inverse undo each other's operations.
Verification Process: To verify this property, let's consider what happens when we apply $f$ to $x$ . We get some output $y$ , such that $y = f(x)$ . Now, if we apply the inverse function $f^{-1}$ to $y$ , we should get back our original input $x$ , because the inverse function undoes the action of the function. So, $f^{-1}(f(x)) = x$ .
Conclusion: Similarly, if we start with some value $x$ and apply the inverse function $f^{-1}$ , we get some output $y$ such that $y = f^{-1}(x)$ . Now, if we apply the function $f$ to $y$ , we should get back our original input $x$ , because the function undoes the action of the inverse function. So, $f(f^{-1}(x)) = x$ .
Conclusion: Similarly, if we start with some value $x$ and apply the inverse function $f^{-1}$ , we get some output $y$ such that $y = f^{-1}(x)$ . Now, if we apply the function $f$ to $y$ , we should get back our original input $x$ , because the function undoes the action of the inverse function. So, $f(f^{-1}(x)) = x$ .Since both $f^{-1}(f(x))$ and $f(f^{-1}(x))$ result in $x$ , we can conclude that the property $f^{-1}$ $1$ holds true for any function $f$ and its inverse $f^{-1}$ , provided that $f$ is invertible and the domain and codomain are appropriately defined so that the compositions are valid.