Answer
$\frac{g}{f}(x) = \frac{1}{2x - x^2}$
$\text{Domain: all real numbers except } 0$ and $-2$
Work Step by Step
This exercise asks us to divide one function by another. Let's write out the problem:
$\frac{g}{f}(x) = g(x) \div f(x) = \frac{\frac{1}{x}}{2 - x}$
When we divide by a fraction, we multiply by its reciprocal:
$\frac{g}{f}(x) = g(x) \div f(x) = (\frac{1}{x})(\frac{1}{2 - x})$
Multiply:
$\frac{g}{f}(x) = g(x) \div f(x) = \frac{1}{2x - x^2}$
When we find the domain, we want to find which values of $x$ will cause the function to become undefined; in other words, we want to find any restrictions for $x$, so we set the denominator equal to $0$ and solve:
$2x - x^2 = 0$
Rewrite in terms of decreasing powers:
$-x^2 + 2x = 0$
Factor out any common terms:
$-x(2 + x) = 0$
Set the first factor equal to $0$:
$-x = 0$
Divide both sides by $-1$ to solve for $x$:
$x = 0$
Set the second factor equal to $0$:
$2 + x = 0$
Subtract $2$ from each side of the equation:
$x = -2$
In this exercise, $x$ can be any real number except for $0$ or $-2$.
