How do you find $f^{- 1} (x)$ $f^-1(x)$ given $f (x) = - \frac{1}{x^{2}}$ $f(x)=-1/x^2$ ?

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How do you find $f^{- 1} (x)$ $f^-1(x)$ given $f (x) = - \frac{1}{x^{2}}$ $f(x)=-1/x^2$ ?

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Answer 1 · 2018-04-01T04:44:08

${f (x)}^{- 1} = \pm \sqrt{- \frac{1}{x}}$ $f(x)^-1=+-sqrt(-1/x)$

Explanation:

You substitute the x values for the y values

$x = - \frac{1}{y^{2}}$ $x=-1/y^2$

Then we rearrange for y

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

$y^{2} = - \frac{1}{x}$ $y^2=-1/x$

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

Such a function doesn't exist as you can't have a negative root on the $RR$ plane. Also it fails the function test as you have two x values corresponding to 1 y value.

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How do you find $f^{- 1} (x)$ $f^-1(x)$ given $f (x) = - \frac{1}{x^{2}}$ $f(x)=-1/x^2$ ?

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How do you find $f^{- 1} (x)$ $f^-1(x)$ given $f (x) = - \frac{1}{x^{2}}$ $f(x)=-1/x^2$ ?