Let $f, g: R \rightarrow R$ be defined,respectively,by $f(x) = x + 1$ and $g(x) = 2x - 3$. Find $f+g$,$f-g$,and $\frac{f}{g}$.
$f, g: R \rightarrow R$ are defined as $f(x) = x + 1$ and $g(x) = 2x - 3$.
$(f+g)(x) = f(x) + g(x) = (x + 1) + (2x - 3) = 3x - 2$.
$(f-g)(x) = f(x) - g(x) = (x + 1) - (2x - 3) = x + 1 - 2x + 3 = -x + 4$.
$\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$,where $g(x) \neq 0$.
$\left(\frac{f}{g}\right)(x) = \frac{x + 1}{2x - 3}$,where $2x - 3 \neq 0$,which implies $x \neq \frac{3}{2}$.
$(f+g)(x) = f(x) + g(x) = (x + 1) + (2x - 3) = 3x - 2$.
$(f-g)(x) = f(x) - g(x) = (x + 1) - (2x - 3) = x + 1 - 2x + 3 = -x + 4$.
$\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$,where $g(x) \neq 0$.
$\left(\frac{f}{g}\right)(x) = \frac{x + 1}{2x - 3}$,where $2x - 3 \neq 0$,which implies $x \neq \frac{3}{2}$.
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