We have 2x^2 + x - 3 = 2x^2 - 2x + 3x - 3 = 2x ( x - 1 ) - 3( x - 1 ) = ( x - 1 )( 2x - 3 );
But g(x) is not 0; then, x is not 1;
The domain is R - {1} ;
f(x) / g(x) = ( x - 1 )( 2x - 3 ) / (x - 1 ) = 2x - 3;
Answer:
The domain of [tex]\frac{f(x)}{g(x)}[/tex] is [tex]-\infty \:<x<\infty \:[/tex]
Step-by-step explanation:
We need to find the domain of [tex]\frac{f(x)}{g(x)}[/tex]
The given functions are: [tex]f(x)=2x^{2}+x-3[/tex] and [tex]g(x)=x-1[/tex]
First, we factorize the f(x);
[tex]f(x)=2x^{2}+x-3[/tex]
[tex]f(x)=2x^{2} - 2x +3x -3[/tex]
[tex]f(x)=2x(x- 1) +3(x -1)[/tex]
[tex]f(x)=(2x +3)(x- 1)[/tex]
Now, divide f(x) by g(x),
[tex]\frac{f(x)}{g(x)}[/tex]
[tex]\frac{f(x)}{g(x)}=\frac{(2x +3)(x- 1)}{(x- 1)}[/tex]
simplify,
[tex]\frac{f(x)}{g(x)}=(2x +3)[/tex]
[tex]\mathrm{Domain\:definition}[/tex]
[tex]\mathrm{The\:domain\:of\:a\:function\:is\:the\:set\:of\:input\:or\:argument\:values\:for\:which\:the\:function\:is\:real\:and\:defined}[/tex]
[tex]\mathrm{The\:function\:has\:no\:undefined\:points\:nor\:domain\:constraints.\:\:Therefore,\:\:the\:domain\:is}[/tex]
[tex]-\infty \:<x<\infty \:[/tex]
Therefore, the domain of [tex]\frac{f(x)}{g(x)}[/tex] is [tex]-\infty \:<x<\infty \:[/tex]
