termegolz6

Answered

2022-07-25

Check whether the given functions are inverse of each others:

1) $g(x)=-{x}^{2}-3\phantom{\rule{0ex}{0ex}}f(x)=\sqrt[5]{-x-3}$

2) $g(x)=\frac{4-x}{x}\phantom{\rule{0ex}{0ex}}f(x)=\frac{4}{x}$

3) $f(x)=\frac{-x-1}{x-2}\phantom{\rule{0ex}{0ex}}g(x)=\frac{-2x+1}{-x-1}$

4) $h(x)=\frac{-2-2x}{x}\phantom{\rule{0ex}{0ex}}f(x)=\frac{-2}{x+2}$

1) $g(x)=-{x}^{2}-3\phantom{\rule{0ex}{0ex}}f(x)=\sqrt[5]{-x-3}$

2) $g(x)=\frac{4-x}{x}\phantom{\rule{0ex}{0ex}}f(x)=\frac{4}{x}$

3) $f(x)=\frac{-x-1}{x-2}\phantom{\rule{0ex}{0ex}}g(x)=\frac{-2x+1}{-x-1}$

4) $h(x)=\frac{-2-2x}{x}\phantom{\rule{0ex}{0ex}}f(x)=\frac{-2}{x+2}$

Answer & Explanation

iljovskint

Expert

2022-07-26Added 18 answers

Step 1

1) $g(x)=-{x}^{5}-3\phantom{\rule{0ex}{0ex}}let\text{}g(x)=y\phantom{\rule{0ex}{0ex}}-{x}^{5}-3=y\phantom{\rule{0ex}{0ex}}-{x}^{5}-y+3\phantom{\rule{0ex}{0ex}}{x}^{5}=-y-3\phantom{\rule{0ex}{0ex}}x=(-y-3{)}^{1/5}$

$g(x)=f(x)=\sqrt[5]{-x-3}\phantom{\rule{0ex}{0ex}}7\in 8$

2) $g(x)-\frac{4-x}{x}\phantom{\rule{0ex}{0ex}}f(x)=\frac{4}{x}\phantom{\rule{0ex}{0ex}}let\text{}g(x)=y\Rightarrow \frac{4-x}{x}=y\phantom{\rule{0ex}{0ex}}\frac{4}{x}=y\phantom{\rule{0ex}{0ex}}\frac{4}{2}=y+1\phantom{\rule{0ex}{0ex}}x=\frac{4}{y+1}$

Step 2

3) $f(x)=\frac{-x-1}{x-2}\phantom{\rule{0ex}{0ex}}let\text{}y=\frac{-x-1}{x-2}\phantom{\rule{0ex}{0ex}}yx-2y=-x-1\phantom{\rule{0ex}{0ex}}yx+x=-1+2y\phantom{\rule{0ex}{0ex}}x(y+1)=-1+2y\phantom{\rule{0ex}{0ex}}x=\frac{-1+2y}{y+1}\phantom{\rule{0ex}{0ex}}f(x)\ne g(x)$

4) $h(x)=\frac{-2-2x}{x}\phantom{\rule{0ex}{0ex}}let\text{}y=\frac{-2-2x}{x}\phantom{\rule{0ex}{0ex}}y=\frac{-2}{x}-2\phantom{\rule{0ex}{0ex}}y+2=\frac{-2}{x}\phantom{\rule{0ex}{0ex}}x=\frac{-2}{y+2}\phantom{\rule{0ex}{0ex}}=7(x)\phantom{\rule{0ex}{0ex}}$

1) $g(x)=-{x}^{5}-3\phantom{\rule{0ex}{0ex}}let\text{}g(x)=y\phantom{\rule{0ex}{0ex}}-{x}^{5}-3=y\phantom{\rule{0ex}{0ex}}-{x}^{5}-y+3\phantom{\rule{0ex}{0ex}}{x}^{5}=-y-3\phantom{\rule{0ex}{0ex}}x=(-y-3{)}^{1/5}$

$g(x)=f(x)=\sqrt[5]{-x-3}\phantom{\rule{0ex}{0ex}}7\in 8$

2) $g(x)-\frac{4-x}{x}\phantom{\rule{0ex}{0ex}}f(x)=\frac{4}{x}\phantom{\rule{0ex}{0ex}}let\text{}g(x)=y\Rightarrow \frac{4-x}{x}=y\phantom{\rule{0ex}{0ex}}\frac{4}{x}=y\phantom{\rule{0ex}{0ex}}\frac{4}{2}=y+1\phantom{\rule{0ex}{0ex}}x=\frac{4}{y+1}$

Step 2

3) $f(x)=\frac{-x-1}{x-2}\phantom{\rule{0ex}{0ex}}let\text{}y=\frac{-x-1}{x-2}\phantom{\rule{0ex}{0ex}}yx-2y=-x-1\phantom{\rule{0ex}{0ex}}yx+x=-1+2y\phantom{\rule{0ex}{0ex}}x(y+1)=-1+2y\phantom{\rule{0ex}{0ex}}x=\frac{-1+2y}{y+1}\phantom{\rule{0ex}{0ex}}f(x)\ne g(x)$

4) $h(x)=\frac{-2-2x}{x}\phantom{\rule{0ex}{0ex}}let\text{}y=\frac{-2-2x}{x}\phantom{\rule{0ex}{0ex}}y=\frac{-2}{x}-2\phantom{\rule{0ex}{0ex}}y+2=\frac{-2}{x}\phantom{\rule{0ex}{0ex}}x=\frac{-2}{y+2}\phantom{\rule{0ex}{0ex}}=7(x)\phantom{\rule{0ex}{0ex}}$

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