Given `g(x)= 2x^2-4x-5,` find the following, `g^(-1)(-2).`

Saturday, December 25, 2010

Given $\displaystyle{g{{\left({x}\right)}}}={2}{{x}}^{{2}}-{4}{x}-{5},$ find the following, $\displaystyle{{g}}^{{-{1}}}{\left(-{2}\right)}.$

Hello!

Actually, this function $\displaystyle{g{}}$ has no (one-valued) inverse function. For some y's there is no such $\displaystyle{x}$ that $\displaystyle{g{{\left({x}\right)}}}={y},$ for some there are two such x's (two solutions of a quadratic equation). But we can consider two-valued functions, too.

The function $\displaystyle{a}{{x}}^{{2}}+{b}{x}+{c}$ with a positive $\displaystyle{a}$ has its minimum at $\displaystyle{x}_{{0}}=\frac{{-{b}}}{{{2}{a}}}.$ For our function $\displaystyle{g{}}$ it is $\displaystyle-\frac{{-{4}}}{{4}}={1},$ the minimum value is $\displaystyle{g{{\left({1}\right)}}}={2}-{4}-{5}=-{7}.$ So there are two points $\displaystyle{x}_{{1}}$ and $\displaystyle{x}_{{2}}$ such that $\displaystyle{g{{\left({x}\right)}}}=-{2}.$ To find them, we have to solve the quadratic equation

$\displaystyle{2}{{x}}^{{2}}-{4}{x}-{5}=-{2},$ or $\displaystyle{2}{{x}}^{{2}}-{4}{x}-{3}={0}.$

The solutions are $\displaystyle{x}_{{{1},{2}}}=\frac{{{2}\pm\sqrt{{{{2}}^{{2}}+{2}\cdot{3}}}}}{{2}}=\frac{{{2}\pm\sqrt{{{10}}}}}{{2}}={1}\pm\sqrt{{\frac{{5}}{{2}}}}.$

The final answer depends on the additional conditions. One may state that there are no $\displaystyle{{g}}^{{-{1}}}{\left(-{2}\right)},$ that there are two values, $\displaystyle{1}-\sqrt{{\frac{{5}}{{2}}}}$ and $\displaystyle{1}+\sqrt{{\frac{{5}}{{2}}}},$ or choose one of them if there are some constraints on the domain of $\displaystyle{g{.}}$