Question

Solve:
f(x)= 2x^2-x-6 and g(x) = x^2+3x-10 What is (f/g) (x) Then find the domain of (f/g) (x) Thanks

Answer 1

To solve for \((f/g)(x)\), we start by defining the functions \(f(x)\) and \(g(x)\):

\[
f(x) = 2x^2 - x - 6
\]
\[
g(x) = x^2 + 3x - 10
\]

Now, to find \((f/g)(x)\), we compute:

\[
\frac{f(x)}{g(x)} = \frac{2x^2 - x - 6}{x^2 + 3x - 10}
\]

Next, we will simplify this expression if possible, by factoring both the numerator and the denominator.

**Step 1: Factor \(f(x)\)**

We want to factor \(f(x) = 2x^2 - x - 6\). To factor, we look for two numbers that multiply to \(2 \times -6 = -12\) and add to \(-1\). The numbers \(-4\) and \(3\) work.

Now we rewrite:

\[
2x^2 - 4x + 3x - 6 = 2x(x - 2) + 3(x - 2) = (2x + 3)(x - 2)
\]

Thus,

\[
f(x) = (2x + 3)(x - 2)
\]

**Step 2: Factor \(g(x)\)**

Now we factor \(g(x) = x^2 + 3x - 10\). We need numbers that multiply to \(-10\) and add to \(3\). The numbers \(5\) and \(-2\) work.

So we write:

\[
g(x) = (x + 5)(x - 2)
\]

**Step 3: Simplify \((f/g)(x)\)**

Now we can rewrite \((f/g)(x)\):

\[
\frac{f(x)}{g(x)} = \frac{(2x + 3)(x - 2)}{(x + 5)(x - 2)}
\]

We see that \((x - 2)\) cancels out.

Thus, we have:

\[
\frac{f(x)}{g(x)} = \frac{2x + 3}{x + 5} \quad (x \neq 2)
\]

**Step 4: Find the domain of \((f/g)(x)\)**

The domain of \(\frac{f(x)}{g(x)}\) is restricted by the values that make \(g(x) = 0\). Setting \(g(x)\) to zero:

\[
g(x) = x^2 + 3x - 10 = 0
\]

We can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 + 40}}{2} = \frac{-3 \pm \sqrt{49}}{2} = \frac{-3 \pm 7}{2}
\]

Calculating the two possible values:

1. \(x = \frac{4}{2} = 2\)
2. \(x = \frac{-10}{2} = -5\)

Thus, the function \(g(x)\) is zero at \(x = 2\) and \(x = -5\). Therefore, \((f/g)(x)\) is undefined at these points.

**Final Domain:**

The domain of \((f/g)(x)\) is:

\[
\text{Domain} = \{ x \in \mathbb{R} \, | \, x \neq 2 \, \text{and} \, x \neq -5 \}
\]

So, in interval notation, the domain is:

\[
(-\infty, -5) \cup (-5, 2) \cup (2, \infty)
\]