Question

Q1 Given the following:
f(x)=2x^2-x-6
g(x)=x^2+3x-10

What is (f/g)(x)? Then, find the domain of (f/g)(x)

Answer 1

To find \((f/g)(x)\), we'll first express it as:

\[
(f/g)(x) = \frac{f(x)}{g(x)}
\]

Given the functions:

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

We need to compute \((f/g)(x)\):

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

### Step 1: Factoring the Functions

**Factoring \(f(x)\)**:

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

Now we can rewrite \(f(x)\) as:

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

So,

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

**Factoring \(g(x)\)**:

To factor \(g(x) = x^2 + 3x - 10\), we look for two numbers that multiply to \(-10\) and add to \(3\). The numbers are \(5\) and \(-2\).

So we can factor \(g(x)\) as:

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

### Step 2: Now Substitute the Factored Forms

We have:

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

### Step 3: Simplifying \((f/g)(x)\)

Notice that \((x - 2)\) appears in both the numerator and the denominator, so we can cancel it out, but we must note that \(x \neq 2\) because division by zero is undefined.

Thus, we simplify:

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

### Step 4: Finding the Domain of \((f/g)(x)\)

The domain of \((f/g)(x)\) includes all real numbers except where the denominator is zero. Therefore, we set \(g(x) = 0\):

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

Factoring gives:

\[
(x + 5)(x - 2) = 0
\]

Setting each factor to zero:

\[
x + 5 = 0 \quad \Rightarrow \quad x = -5
\]
\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]

So the values that \(x\) cannot take are \(-5\) and \(2\).

### Conclusion

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

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

In interval notation, this is:

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