Solving for zeroes and non-domain points: For each of the following expressions, find all values of x for which g(x) is zero or does not exist.

$$\begin{gathered} 1.g\left(x\right)=\frac{3x^2-x-2}{x^4+2x^2-3}. \\ 2.g\left(x\right)=\frac{1}{x-2}-\frac{3x+1}{x+2}. \\ 3.g\left(x\right)=\frac{\sin x}{\cos x}. \\ 4.g\left(x\right)=\frac{e^{3x}(x-1)}{\ln x}. \end{gathered}$$

Given the following functions:

$$\begin{gathered} 1.g\left(x\right)=\frac{3x^2-x-2}{x^4+2x^2-3}. \\ 2.g\left(x\right)=\frac{1}{x-2}-\frac{3x+1}{x+2}. \\ 3.g\left(x\right)=\frac{\sin x}{\cos x}. \\ 4.g\left(x\right)=\frac{e^{3x}(x-1)}{\ln x}. \end{gathered}$$

$$\begin{gathered} g\left(x\right)=0whennumeratoriszeroi.e., \\ 3x^2-x-2=0 \\ x=1,\frac{-2}{3}. \\ andg\left(x\right)\hspace{0.17em}doesnotexistwhendenominatoriszeroso, \\ {x}^4+2x^2-3=0 \\ {x}^2=1,-3. \\ Since,squarecannotbenegativeso \\ {x}^2=1, \\ andx=1,-1. \\ Combiningallthevaluesofxwegetx=1,-1,-\frac{2}{3}. \\ Sincex=1isacommonrootthereforeitwillbecancelledandweget, \\ x=-1,-\frac{2}{3}. \end{gathered}$$

$$\begin{gathered} g\left(x\right)=0whennumeratoriszeroi.e., \\ 1cannotbeequaltozero. \\ Now,3x+1=0 \\ givesx=-\frac{1}{3}. \\ andg\left(x\right)\hspace{0.17em}doesnotexistwhendenominatoriszeroso, \\ x-2=0orx+2=0 \\ weget, \\ x=2,-2. \\ Combiningallthevaluesofxwegetx=2,-2,-\frac{1}{3}. \end{gathered}$$

$$\begin{gathered} g\left(x\right)=0whennumeratoriszeroi.e., \\ \sin x=0 \\ x=n\mathrm{\pi },\mathrm{where}\mathrm{n}\in \mathrm{I}. \\ andg\left(x\right)\hspace{0.17em}doesnotexistwhendenominatoriszeroso, \\ \cos x=0 \\ x=\left(2n+1\right)\frac{\mathrm{\pi }}{2},n\in I. \\ Combiningallthevaluesofxwegetx=n\mathrm{\pi },\left(2\mathrm{n}+1\right)\frac{\mathrm{\pi }}{2}\mathrm{where}\mathrm{n}\in \mathrm{I}. \end{gathered}$$

$$\begin{gathered} g\left(x\right)=0whennumeratoriszeroi.e., \\ {e}^{3x}(x-1)=0 \\ Since,e^{3x}cannotbezeroso,x-1=0 \\ Andx=1. \\ andg\left(x\right)\hspace{0.17em}doesnotexistwhendenominatoriszeroso, \\ \ln x=0. \\ x=1. \\ Combiningallthevaluesofxwegetx=1. \\ Sincex=1isacommonrootthereforeitwillbecancelledandweget, \\ nothingfromherebutifweseeoutofthedomainallthenegativevaluesofx \\ andx=0willcausethefunctiong\left(x\right)notexist. \\ Therefore,x\in (-\infty ,0]. \end{gathered}$$