What is the inverse of $h (x) = - \frac{4}{x^{2}}$ $h(x)=-4/x^2$ ?

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The furthest I got on this question was $y = \sqrt{- \frac{1}{4} x}$ $y=sqrt(-1/4x$

But I don't think that's right

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Answer 1 · 2016-10-23T14:43:29

Do you want + or -?

Exchange x and y.

$x = - \frac{4}{y^{2}}$ $x = -4/y^2$

Multiply by $y^{2}$ $y^2$

$x y^{2} = - 4$ $xy^2 = -4$

Divide by x, then take the root.

$y^{2} = - \frac{4}{x}$ $y^2 = -4/x$

$y = \pm \sqrt{- \frac{4}{x}} = h^{- 1} (x)$ $y = ± sqrt{-4/x} = h^-1(x)$