Inverses of Quadratic Functions.

Question

The domain of $g(x) = x-x^2$ is such that $g^{-1}(x)$ exists. Explain why $x \geq 1$ is a suitable domain for $g(x)$? I'm guessing it has something to do with negative values not giving possible outputs for $g^{-1}(x)$ but I don't know how to explain it technically.

Answer 1

Actually inverse $g^{-1}$ exists also for $x\ge\frac12$

To a greater extent $g(x)$ is invertible for $x\ge 1$

To be invertible any horizontal line must intersect in only one point the graph of the function, which is equivalent to ask that the function is injective.

To complete the answer it must be said that also $x\le \frac12$ is a suitable domain where $g(x)$ can be inverted

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Answer 2

Let $y=x-x^2$.

Thus, $$x-x^2=x(1-x)\leq\left(\frac{x+1-x}{2}\right)^2=\frac{1}{4},$$ which says $y\leq\frac{1}{4}$ and $$x^2-x+y=0,$$ which gives $$x=\frac{1+\sqrt{1-4y}}{2}$$ or $$x=\frac{1-\sqrt{1-4y}}{2}.$$ Now we see that for $g$ with domain $x\geq\frac{1}{2}$ we have $$g^{-1}(x)=\frac{1+\sqrt{1-4x}}{2}$$ and for $g$ with domain $x\leq\frac{1}{2}$ we have $$g^{-1}(x)=\frac{1-\sqrt{1-4x}}{2}.$$