Need help with the following: if `f(x) = 2x-3;` `g(x) = 1/(x+1),` find `(f@g)^(-1)(x).`

Saturday, September 12, 2009

Need help with the following: if $\displaystyle{f{{\left({x}\right)}}}={2}{x}-{3};$ $\displaystyle{g{{\left({x}\right)}}}=\frac{{1}}{{{x}+{1}}},$ find $\displaystyle{{\left({f{\circ}}{g}\right)}}^{{-{1}}}{\left({x}\right)}.$

Hello!

First found the composition:

$\displaystyle{\left({f{\circ}}{g}\right)}{\left({x}\right)}={f{{\left({g{{\left({x}\right)}}}\right)}}}={f{{\left(\frac{{1}}{{{x}+{1}}}\right)}}}={2}\cdot\frac{{1}}{{{x}+{1}}}-{3}=-\frac{{{3}{x}+{1}}}{{{x}+{1}}}.$

This function is defined everywhere except $\displaystyle{x}=-{1}.$

To find the inverse function, we have to solve for $\displaystyle{x}$ the equation $\displaystyle{\left({f{\circ}}{g}\right)}{\left({x}\right)}={y}:$

$\displaystyle-\frac{{{3}{x}+{1}}}{{{x}+{1}}}={y},$ or $\displaystyle\frac{{{3}{x}+{1}}}{{{x}+{1}}}=-{y}.$

Multiply both sides by $\displaystyle{\left({x}+{1}\right)}$ and obtain

$\displaystyle{3}{x}+{1}=-{x}{y}-{y}.$

Move the terms with $\displaystyle{x}$ to the left and without $\displaystyle{x}$ to the right:

$\displaystyle{x}{\left({y}+{3}\right)}=-{y}-{1}=-{\left({y}+{1}\right)},$

so $\displaystyle{x}=-\frac{{{y}+{1}}}{{{y}+{3}}}$ (of course $\displaystyle{y}\ne-{3}$ ).

Thus the answer is: the function $\displaystyle{{\left({f{\circ}}{g}\right)}}^{{-{1}}}{\left({y}\right)}$ exists for all $\displaystyle{y}\ne-{3}$ and is equal to $\displaystyle-\frac{{{y}+{1}}}{{{y}+{3}}}.$

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Saturday, September 12, 2009

Need help with the following: if $\displaystyle{f{{\left({x}\right)}}}={2}{x}-{3};$ $\displaystyle{g{{\left({x}\right)}}}=\frac{{1}}{{{x}+{1}}},$ find $\displaystyle{{\left({f{\circ}}{g}\right)}}^{{-{1}}}{\left({x}\right)}.$

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What was the device called which Faber had given Montag in order to communicate with him?

Saturday, September 12, 2009

Need help with the following: if find

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What was the device called which Faber had given Montag in order to communicate with him?

Need help with the following: if $\displaystyle{f{{\left({x}\right)}}}={2}{x}-{3};$ $\displaystyle{g{{\left({x}\right)}}}=\frac{{1}}{{{x}+{1}}},$ find $\displaystyle{{\left({f{\circ}}{g}\right)}}^{{-{1}}}{\left({x}\right)}.$