It's been a long while since my school days and linear algebra. I'd like to find the function, if one exists, that transforms $$y = 2x^2 - 1$$ to $$y = x.$$ Any help is appreciated.
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$\sqrt{y+1\over2}$ – Ivan Neretin Feb 11 '23 at 23:32
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Let $f(x)=2x^2-1$. You are looking for the inverse function of $f$, denoted as $f^{-1}$, that satisfies $$f^{-1}(f(x))=f(f^{-1}(x))=x.$$
To find it, we just need to represent the input $x$ in terms of the output $y$: $$ \begin{align} y&=2x^2-1 \\ 2x^2&=y+1 \\ x&=\sqrt \frac{y+1}{2}. \end{align} $$ The inverse is therefore $$ f^{-1}(y)=\sqrt \frac{y+1}{2}. $$
Scene
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Thanks, all. @Scene answered my question for those particular functions. More generally, if I might, I'm looking for h(f) to transform quadratic f(x) to linear g(x). Say, for example, f(x) = 2(x^2) - 1 and g(x) = 3x + 4. – user20730 Feb 12 '23 at 00:32
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I am not sure I understand your question. Fwiw, for each real number $y>-1,$ there are two real numbers $x$ such that this $y$ is equal to $2x^2-1:$ $x=\sqrt{y+1\over2}$ and $x=-\sqrt{y+1\over2}.$
Anne Bauval
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