Given that \( f(x)=x^{2}-9 x \) and \( g(x)=x+12 \), find (a) \( f+g=\square \) and its domain is \( \square \) and its domain is \( \square \) and its domain is \( \square \) (b) \( f-g=\square \) and its domain is \( x \neq \square \) (c) \( f g=\square \) (d) \( \frac{f}{g}=\square \)

Let's solve each part step by step. ### Given Functions: - \( f(x) = x^{2} - 9x \) - \( g(x) = x + 12 \) --- ### (a) \( f + g \) **Calculation:** \[ f(x) + g(x) = (x^{2} - 9x) + (x + 12) = x^{2} - 8x + 12 \] **Expression:** \[ f + g = x^{2} - 8x + 12 \] **Domain:** Both \( f(x) \) and \( g(x) \) are polynomials, which are defined for all real numbers. \[ \text{Domain of } f + g: \mathbb{R} \] --- ### (b) \( f - g \) **Calculation:** \[ f(x) - g(x) = (x^{2} - 9x) - (x + 12) = x^{2} - 10x - 12 \] **Expression:** \[ f - g = x^{2} - 10x - 12 \] **Domain:** Both \( f(x) \) and \( g(x) \) are polynomials, which are defined for all real numbers. \[ \text{Domain of } f - g: \mathbb{R} \] --- ### (c) \( f \cdot g \) **Calculation:** \[ f(x) \cdot g(x) = (x^{2} - 9x)(x + 12) \] \[ = x^{3} + 12x^{2} - 9x^{2} - 108x \] \[ = x^{3} + 3x^{2} - 108x \] **Expression:** \[ f \cdot g = x^{3} + 3x^{2} - 108x \] **Domain:** Both \( f(x) \) and \( g(x) \) are polynomials, which are defined for all real numbers. \[ \text{Domain of } f \cdot g: \mathbb{R} \] --- ### (d) \( \frac{f}{g} \) **Calculation:** \[ \frac{f(x)}{g(x)} = \frac{x^{2} - 9x}{x + 12} \] **Expression:** \[ \frac{f}{g} = \frac{x^{2} - 9x}{x + 12} \] **Domain:** The denominator \( g(x) = x + 12 \) must not be zero. \[ x + 12 \neq 0 \implies x \neq -12 \] \[ \text{Domain of } \frac{f}{g}: x \in \mathbb{R}, x \neq -12 \] --- ### Summary of Answers: **(a)** \[ f + g = x^{2} - 8x + 12 \] \[ \text{Domain}: \mathbb{R} \] **(b)** \[ f - g = x^{2} - 10x - 12 \] \[ \text{Domain}: \mathbb{R} \] **(c)** \[ f \cdot g = x^{3} + 3x^{2} - 108x \] \[ \text{Domain}: \mathbb{R} \] **(d)** \[ \frac{f}{g} = \frac{x^{2} - 9x}{x + 12} \] \[ \text{Domain}: x \neq -12 \]