Answer:
Step-by-step explanation:
Remember:
(fog)(x)= g(f(x))
a)
[tex](fog)(x)=2x-\dfrac{1}{x} =g(f(x))\\\\g(x)=5x+2\\\\Calculate\ g^{-1}(x)\\\\y=5x+2\\\\x=5y+2\ exchanging\ x\ and\ y\\\\y=\dfrac{x-2}{5} \\\\\boxed{g^{-1}(x)=\dfrac{x-2}{5}}\\\\[/tex]
[tex][(f\ o\ g)\ o\ g^{-1}](x)=[f\ o\ (g\ o \ g^{-1})](x)=f(x)\\\\g^{-1}(g(f(x))=g^{-1}(2x-\dfrac{1}{x} )=\dfrac{2x-\dfrac{1}{x}-2}{5} =\dfrac{2x}{5} -\dfrac{1}{5x}-\dfrac{2}{5} \\\\\\\boxed{f(x)=\dfrac{2x}{5} -\dfrac{1}{5x}-\dfrac{2}{5} }\\\\[/tex]
b)
[tex]g^{-1}(x)+3=f(g(x))\\\\\dfrac{x-2}{5} +3=f(g(x))\\\\\dfrac{x-2}{5} +3=f(5x+2)\\\\\dfrac{x-2}{5} +3=\dfrac{2(5x+2)}{5} -\dfrac{1}{5(5x+2)}-\dfrac{2}{5} \\\\\dfrac{x}{5} -\dfrac{2}{5} +\dfrac{15}{5}=\dfrac{10x}{5} +\dfrac{4}{5} -\dfrac{1}{5(5x+2)}-\dfrac{2}{5} \\\\9x-\dfrac{1}{5x+2}=11\\\\x\neq -\dfrac{2}{5} \\\\[/tex]
[tex]45x^2-37x-23=0\\\Delta=37^2+4*23*45=5509\\\\Sol=\{ \dfrac{37-\sqrt{5509} }{90} ,\dfrac{37+\sqrt{5509} }{90} \}[/tex]
