Let f(x)=7x+3 and g(x)=2x-7 Find (f+g)(x),(f-g)(x),(fg)(x) and ((f)/(g))(x) Give the domain of each. (f+g)(x)=9x-4 (Simplify your answer.) (f-g)(x)= square (Simplify your answer.)

To find $(f+g)(x)$, we need to add the functions $f(x)$ and $g(x)$ together.<br /><br />$(f+g)(x) = f(x) + g(x)$<br /><br />Substituting the given functions, we have:<br /><br />$(f+g)(x) = (7x+3) + (2x-7)$<br /><br />Combining like terms, we get:<br /><br />$(f+g)(x) = 9x - 4$<br /><br />Therefore, $(f+g)(x) = 9x - 4$.<br /><br />Next, let's find $(f-g)(x)$.<br /><br />$(f-g)(x) = f(x) - g(x)$<br /><br />Substituting the given functions, we have:<br /><br />$(f-g)(x) = (7x+3) - (2x-7)$<br /><br />Combining like terms, we get:<br /><br />$(f-g)(x) = 5x + 10$<br /><br />Therefore, $(f-g)(x) = 5x + 10$.<br /><br />Now, let's find $(fg)(x)$.<br /><br />$(fg)(x) = f(x) \cdot g(x)$<br /><br />Substituting the given functions, we have:<br /><br />$(fg)(x) = (7x+3) \cdot (2x-7)$<br /><br />Expanding the expression, we get:<br /><br />$(fg)(x) = 14x^2 - 49x + 6x - 21$<br /><br />Combining like terms, we get:<br /><br />$(fg)(x) = 14x^2 - 43x - 21$<br /><br />Therefore, $(fg)(x) = 14x^2 - 43x - 21$.<br /><br />Finally, let's find $(\frac{f}{g})(x)$.<br /><br />$(\frac{f}{g})(x) = \frac{f(x)}{g(x)}$<br /><br />Substituting the given functions, we have:<br /><br />$(\frac{f}{g})(x) = \frac{7x+3}{2x-7}$<br /><br />Therefore, $(\frac{f}{g})(x) = \frac{7x+3}{2x-7}$.<br /><br />The domain of each function is the set of all real numbers except for any values that make the denominator zero. In this case, the denominator is $2x-7$, so the domain is all real numbers except $x = \frac{7}{2}$.<br /><br />In summary:<br />$(f+g)(x) = 9x - 4$<br />$(f-g)(x) = 5x + 10$<br />$(fg)(x) = 14x^2 - 43x - 21$<br />$(\frac{f}{g})(x) = \frac{7x+3}{2x-7}$<br /><br />The domain of each function is all real numbers except $x = \frac{7}{2}$.