Let's find the answers for each part step by step.
Given:
- \( f(x) = x^2 - 3x \)
- \( g(x) = 6 + x \)
Part 1: \( (f + g)(x) \)
To find \( (f + g)(x) \): \[ (f + g)(x) = f(x) + g(x) \] \[ = (x^2 - 3x) + (6 + x) \] Now, combine like terms: \[ = x^2 - 3x + x + 6 \] \[ = x^2 - 2x + 6 \]
So, \[ \text{(a) } (f + g)(x) = x^2 - 2x + 6 \]
Part 2: \( (f - g)(x) \)
To find \( (f - g)(x) \): \[ (f - g)(x) = f(x) - g(x) \] \[ = (x^2 - 3x) - (6 + x) \] Now, combine like terms: \[ = x^2 - 3x - 6 - x \] \[ = x^2 - 4x - 6 \]
So, \[ \text{(b) } (f - g)(x) = x^2 - 4x - 6 \]
Part 3: \( (f \times g)(x) \)
To find \( (f \times g)(x) \): \[ (f \times g)(x) = f(x) \cdot g(x) \] \[ = (x^2 - 3x)(6 + x) \] Distributing this: \[ = x^2(6 + x) - 3x(6 + x) \] \[ = 6x^2 + x^3 - 18x - 3x^2 \] Combining like terms: \[ = x^3 + 3x^2 - 18x \]
So, \[ \text{(c) } (f \times g)(x) = x^3 + 3x^2 - 18x \]
Part 4: \( \frac{f}{g}(x) \)
To find \( \left(\frac{f}{g}\right)(x) \): \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} \] \[ = \frac{x^2 - 3x}{6 + x} \]
This is already in its simplified form.
So, \[ \text{(d) } \left( \frac{f}{g} \right)(x) = \frac{x^2 - 3x}{6 + x} \]
Part 5: Domain of \( \frac{f}{g} \)
To find the domain of \( \frac{f}{g} \), we need to determine where \( g(x) \neq 0 \): \[ g(x) = 6 + x \neq 0 \] Hence, \[ x \neq -6 \]
In interval notation, the domain is: \[ (-\infty, -6) \cup (-6, \infty) \]
Summary of Answers
(a) \( (f + g)(x) = x^2 - 2x + 6 \)
(b) \( (f - g)(x) = x^2 - 4x - 6 \)
(c) \( (f \times g)(x) = x^3 + 3x^2 - 18x \)
(d) \( \left( \frac{f}{g} \right)(x) = \frac{x^2 - 3x}{6 + x} \)
(e) The domain of \( \frac{f}{g} \) is \( (-\infty, -6) \cup (-6, \infty) \)