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

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

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Answer 1 · 2016-03-23T22:55:43

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

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

Explanation:

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

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

$y = x^{2} - 1$ $y = x^2-1$

Add $1$ $1$ to both sides to get:

$y + 1 = x^{2}$ $y + 1 = x^2$

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

$x = \pm \sqrt{y + 1}$ $x = +-sqrt(y+1)$

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

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

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

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

1 Answer

Explanation:

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