How do you find the inverse of $f(x)=x^2 - 1$ and is it a function?

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Answer 1 · 2016-03-23T22:55:43

Let $y = f(x)$ and solve for $x$ .

We find that there is no inverse function unless the domain of $f(x)$ is restricted.

Suppose $f(x) = x^2-1$

To attempt to find an inverse function, let $y = f(x)$ and solve for $x$ in terms of $y$ ...

$y = x^2-1$

Add $1$ to both sides to get:

$y + 1 = x^2$

Transpose and take square root of both sides, allowing for either sign:

$x = +-sqrt(y+1)$

This does not determine a unique value for $x$ in terms of $y$ . So there is no inverse function, unless we restrict the domain of $f(x)$ .

For example, if we specify an explicit domain $[0, oo)$ for $f(x)$ , then $f^(-1)(y) = sqrt(y+1)$

Alternatively, we might specify an explicit domain $(-oo, 0]$ for $f(x)$ , then $f^(-1)(y) = -sqrt(y+1)$

How do you find the inverse of f(x)=x^2 - 1f(x)=x^2 - 1 and is it a function?