1. **Problem statement:** Given functions $f(x) = 8x^3$ and $g(x) = -2x^{3/2}$, find $(fg)(x)$ and $\left(\frac{f}{g}\right)(x)$, state their domains, and evaluate both at $x=4$. 2. **Formulas and rules:** - Product of functions: $(fg)(x) = f(x) \cdot g(x)$ - Quotient of functions: $\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$, with $g(x) \neq 0$ - Domain of $f(x) = 8x^3$ is all real numbers since polynomials are defined everywhere. - Domain of $g(x) = -2x^{3/2}$ requires $x \geq 0$ because of the fractional exponent $3/2$ (square root and cube). - Domain of $(fg)(x)$ is intersection of domains of $f$ and $g$, so $x \geq 0$. - Domain of $\left(\frac{f}{g}\right)(x)$ is domain of $f$ and $g$ with $g(x) \neq 0$, so $x \geq 0$ and $g(x) \neq 0$. 3. **Find $(fg)(x)$:** $$ (fg)(x) = f(x) \cdot g(x) = 8x^3 \cdot (-2x^{3/2}) = -16x^{3 + \frac{3}{2}} = -16x^{\frac{6}{2} + \frac{3}{2}} = -16x^{\frac{9}{2}} $$ 4. **Find $\left(\frac{f}{g}\right)(x)$:** $$ \left(\frac{f}{g}\right)(x) = \frac{8x^3}{-2x^{3/2}} = \frac{8}{-2} \cdot \frac{x^3}{x^{3/2}} = -4x^{3 - \frac{3}{2}} = -4x^{\frac{6}{2} - \frac{3}{2}} = -4x^{\frac{3}{2}} $$ 5. **Simplify the quotient step with cancellation:** $$ \frac{8x^3}{-2x^{3/2}} = \frac{\cancel{8} \cdot x^3}{\cancel{2} \cdot x^{3/2}} = 4 \cdot x^{3 - \frac{3}{2}} = 4x^{\frac{3}{2}} $$ Note the negative sign: final is $-4x^{3/2}$. 6. **Domain of $(fg)(x)$:** - Since $g(x)$ involves $x^{3/2}$, $x \geq 0$. - $f(x)$ is defined for all real $x$. - So domain of $(fg)(x)$ is $[0, \infty)$. 7. **Domain of $\left(\frac{f}{g}\right)(x)$:** - Same domain restriction $x \geq 0$. - Also $g(x) \neq 0$ means $-2x^{3/2} \neq 0 \Rightarrow x^{3/2} \neq 0 \Rightarrow x \neq 0$. - So domain is $(0, \infty)$. 8. **Evaluate $(fg)(4)$:** $$ (fg)(4) = -16 \cdot 4^{9/2} = -16 \cdot (4^{4} \cdot 4^{1/2}) = -16 \cdot (256 \cdot 2) = -16 \cdot 512 = -8192 $$ 9. **Evaluate $\left(\frac{f}{g}\right)(4)$:** $$ \left(\frac{f}{g}\right)(4) = -4 \cdot 4^{3/2} = -4 \cdot (4^{1} \cdot 4^{1/2}) = -4 \cdot (4 \cdot 2) = -4 \cdot 8 = -32 $$ **Final answers:** - $(fg)(x) = -16x^{9/2}$ with domain $[0, \infty)$ and $(fg)(4) = -8192$ - $\left(\frac{f}{g}\right)(x) = -4x^{3/2}$ with domain $(0, \infty)$ and $\left(\frac{f}{g}\right)(4) = -32$