Given the following: f(x)=2x2−x−6 g(x)=x2+3x−10 What is (f/g)(x) ? Then, find the domain of (f/g)(x) . (2 points) include a brief summary of how you your answers

1 answer

To find \( (f/g)(x) \), we first express this as:

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

Step 1: Factor \( f(x) \) and \( g(x) \)

Factor \( f(x) \):

To factor \( f(x) = 2x^2 - x - 6 \):

  1. Multiply \( a \) (the coefficient of \( x^2 \)) and \( c \) (the constant term): \( 2 \times -6 = -12 \).
  2. We need two numbers that multiply to -12 and add to -1 (the coefficient of \( x \)): those numbers are \( -4 \) and \( 3 \).

Thus, we can rewrite it: \[ f(x) = 2x^2 - 4x + 3x - 6 \] \[ = 2x(x - 2) + 3(x - 2) \] \[ = (2x + 3)(x - 2) \]

Factor \( g(x) \):

To factor \( g(x) = x^2 + 3x - 10 \):

  1. Find two numbers that multiply to -10 and add to 3: those numbers are \( 5 \) and \( -2 \).

Thus, we can factor it as: \[ g(x) = (x + 5)(x - 2) \]

Step 2: Write \( (f/g)(x) \)

Now substituting the factored forms into \( (f/g)(x) \): \[ \frac{f(x)}{g(x)} = \frac{(2x + 3)(x - 2)}{(x + 5)(x - 2)} \] Notice that \( (x - 2) \) in the numerator and denominator can be cancelled out: \[ = \frac{2x + 3}{x + 5}, \quad x \neq 2 \]

Step 3: Find the Domain of \( (f/g)(x) \)

The domain of \( (f/g)(x) \) is determined by the values of \( x \) for which \( g(x) \neq 0 \), since we cannot divide by zero.

Set \( g(x) = 0 \): \[ x^2 + 3x - 10 = 0 \] Factoring gives: \[ (x + 5)(x - 2) = 0 \] Thus, \[ x + 5 = 0 \quad \text{or} \quad x - 2 = 0 \] This yields: \[ x = -5 \quad \text{or} \quad x = 2 \] These values cannot be included in the domain.

Summary of Results

  1. The simplified form of \( (f/g)(x) \) is: \[ (f/g)(x) = \frac{2x + 3}{x + 5}, \quad x \neq 2 \]

  2. The domain of \( (f/g)(x) \) excludes the values where \( g(x) \) is zero: \[ \text{Domain: } x \in \mathbb{R} \setminus {-5, 2} \]

The process involved factoring both functions, cancelling out common factors, and determining restrictions from the denominator to arrive at the final expression and its domain.