Find the derivative of the function using the definition of derivative. \[ \begin{array}{r} f(x)=3+6 x-8 x^{2} \\ f^{\prime}(x)=\square \end{array} \] State the domain of the function. (Enter your answer using interval notation.) \( \square \) State the domain of its derivative. (Enter your answer using interval notation.) \( \square \)

To find the derivative of the function \( f(x) = 3 + 6x - 8x^{2} \) using the **definition of the derivative**, we'll follow these steps: ### Definition of the Derivative The derivative of a function \( f(x) \) at a point \( x \) is defined as: \[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \] ### Step-by-Step Calculation 1. **Compute \( f(x + h) \):** \[ f(x + h) = 3 + 6(x + h) - 8(x + h)^2 \] Expand the terms: \[ f(x + h) = 3 + 6x + 6h - 8(x^2 + 2xh + h^2) = 3 + 6x + 6h - 8x^2 - 16xh - 8h^2 \] 2. **Find \( f(x + h) - f(x) \):** \[ f(x + h) - f(x) = (3 + 6x + 6h - 8x^2 - 16xh - 8h^2) - (3 + 6x - 8x^2) = 6h - 16xh - 8h^2 \] 3. **Divide by \( h \):** \[ \frac{f(x + h) - f(x)}{h} = \frac{6h - 16xh - 8h^2}{h} = 6 - 16x - 8h \] 4. **Take the limit as \( h \to 0 \):** \[ f'(x) = \lim_{h \to 0} (6 - 16x - 8h) = 6 - 16x \] ### Final Results \[ \begin{array}{r} f(x)=3+6 x-8 x^{2} \\ f^{\prime}(x)=6 - 16x \end{array} \] **Domain of the Original Function \( f(x) \):** Since \( f(x) \) is a polynomial, it is defined for all real numbers. \[ \text{Domain of } f(x): \quad (-\infty, \infty) \] **Domain of the Derivative \( f'(x) \):** Similarly, \( f'(x) = 6 - 16x \) is also a polynomial and is defined for all real numbers. \[ \text{Domain of } f'(x): \quad (-\infty, \infty) \] --- **Summary of Answers:** \[ \begin{array}{r} f'(x) = 6 - 16x \\ \text{Domain of } f(x): \quad (-\infty, \infty) \\ \text{Domain of } f'(x): \quad (-\infty, \infty) \end{array} \]