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The function $ f(x)=\sqrt{x+4} $ has domain $[-4,\infty)$ and range $[0,\infty)$.

Its inverse function $ f^{-1}(x)=x^2-4 $ by itself seems to have domain $\Bbb{R}$ and range $[-4,\infty)$.

However, we learned in school that a function and its inverse always have swapped domain and ranges, so should I say its "actual" domain is $[0,\infty)$?

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The domain is the inputs a function can take. The range is the outputs a function can give.

Taking the inverse switches the two. You are now giving the outputs, and getting back the inputs.

The domain of the inverse is the possible outputs of the original function, which was it's range. The range of the inverse is the possible inputs of the original function, which was it's domain.

You have a special case. The inverse of $f$ is not $x^2 - 4$, but instead the right half of $x^2 - 4$. By looking at the original function's range, you can make sense of this.

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You have to restrict the function's domain as necessary in order for things to work. See, $f^{-1}(x)$ right now is not injective, so it doesn't have a globally defined inverse. Its restriction to $\mathbb{R}_{>0}$, however, is fine. So yes, the domain and image have to swap.

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Think about how the inverse function behaves with respect to the original function. They mirror each other.


EDIT 1 The inverse function's domain and range are related to the original function's domain and range.


EDIT 2 They're opposites(duals). Hope it helps.

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  • $\begingroup$ I would say rather then opposites duals. $\endgroup$ – Q the Platypus Dec 22 '16 at 4:39

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