Let f(x)=4x^2-3x and g(x)=x^2-x+3 Find (f+g)(x),(f-g)(x),(fg)(x) and ((f)/(g))(x) Give the domain of each. (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 />$(f+g)(x) = (4x^2 - 3x) + (x^2 - x + 3)$<br /><br />Combining like terms, we get:<br /><br />$(f+g)(x) = 5x^2 - 4x + 3$<br /><br />Therefore, $(f+g)(x) = 5x^2 - 4x + 3$.<br /><br />The domain of $(f+g)(x)$ is all real numbers, since there are no restrictions on the values of $x$ for which the expression is defined.<br /><br />$(f-g)(x)=$ $\square $ (Simplify your answer.)<br /><br />To find $(f-g)(x)$, we need to subtract the function $g(x)$ from the function $f(x)$.<br /><br />$(f-g)(x) = f(x) - g(x)$<br /><br />Substituting the given functions, we have:<br /><br />$(f-g)(x) = (4x^2 - 3x) - (x^2 - x + 3)$<br /><br />Combining like terms, we get:<br /><br />$(f-g)(x) = 3x^2 - 2x - 3$<br /><br />Therefore, $(f-g)(x) = 3x^2 - 2x - 3$.<br /><br />The domain of $(f-g)(x)$ is all real numbers, since there are no restrictions on the values of $x$ for which the expression is defined.<br /><br />$(fg)(x)=$ $\square $ (Simplify your answer.)<br /><br />To find $(fg)(x)$, we need to multiply the functions $f(x)$ and $g(x)$ together.<br /><br />$(fg)(x) = f(x) \cdot g(x)$<br /><br />Substituting the given functions, we have:<br /><br />$(fg)(x) = (4x^2 - 3x) \cdot (x^2 - x + 3)$<br /><br />Expanding the expression, we get:<br /><br />$(fg)(x) = 4x^4 - 7x^3 + 9x^2 - 3x^3 + 3x^2 - 9x$<br /><br />Combining like terms, we get:<br /><br />$(fg)(x) = 4x^4 - 10x^3 + 12x^2 - 9x$<br /><br />Therefore, $(fg)(x) = 4x^4 - 10x^3 + 12x^2 - 9x$.<br /><br />The domain of $(fg)(x)$ is all real numbers, since there are no restrictions on the values of $x$ for which the expression is defined.<br /><br />$(\frac {f}{g})(x)=$ $\square $ (Simplify your answer.)<br /><br />To find $(\frac {f}{g})(x)$, we need to divide the function $f(x)$ by the function $g(x)$.<br /><br />$(\frac {f}{g})(x) = \frac{f(x)}{g(x)}$<br /><br />Substituting the given functions, we have:<br /><br />$(\frac {f}{g})(x) = \frac{4x^2 - 3x}{x^2 - x + 3}$<br /><br />Therefore, $(\frac {f}{g})(x) = \frac{4x^2 - 3x}{x^2 - x + 3}$.<br /><br />The domain of $(\frac {f}{g})(x)$ is all real numbers except for the values of $x$ that make the denominator equal to zero. In this case, the denominator is $x^2 - x + 3$, which is never equal to zero for any real number $x$. Therefore, the domain of $(\frac {f}{g})(x)$ is all real numbers.