Answer
$\frac{g}{f}(x) = \dfrac{1}{2x + 3}$
$\text{The domain is the set of all real numbers except } -\frac{3}{2}$ and $1$.
Work Step by Step
This exercise asks us to divide one function by another. Let's write out the problem:
$\frac{g}{f}(x) = g(x) \div f(x) = \frac{x - 1}{2x^2 + x - 3}$
Let's factor the denominator by splitting the middle term. We have a quadratic expression, which is in the form $ax^2 + bx + c$. We need to find which factors multiplied together will equal $ac$ but when added together will equal $b$.
In this equation, $ac$ is $-6$ and $b$ is $1$. The factors $3$ and $-2$ will work.
Let's rewrite the equation and split the middle term using these two factors:
$2x^2 + 3x - 2x - 3$
Group the first and third terms and the second and last terms:
$(2x^2 - 2x) + (3x - 3)$
Factor common terms out:
$2x(x - 1) + 3(x - 1)$
Group the factors:
$(2x + 3)(x - 1)$
We rewrite the rational expression using the factored form:
$\frac{g}{f}(x) = g(x) \div f(x) = \frac{x - 1}{(2x + 3)(x - 1)}$
Cancel out common terms in the numerator and denominator:
$\frac{g}{f}(x) = g(x) \div f(x) = \frac{1}{2x + 3}$
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 $x$. We set the denominator equal to $0$ and solve for exercise:
$(2x + 3)(x - 1) = 0$
We set the first factor equal to $0$:
$2x + 3 = 0$
Subtract $3$ from each side of the equation:
$2x = -3$
Divide both sides of the equation by $2$:
$x = -\frac{3}{2}$
Let's set the second factor equal to $0$:
$x - 1= 0$
Add $1$ to each side of the equation:
$x = 1$
In this exercise, $x$ can be any real number except for $-\frac{3}{2}$ and $1$.
