1. **State the problem:** Find the average value of the function $g(x) = 1 - 6x^2$ on the interval $[-1, 3]$. 2. **Formula for average value:** The average value of a function $f(x)$ on $[a,b]$ is given by $$\text{Average value} = \frac{1}{b-a} \int_a^b f(x) \, dx$$ 3. **Apply the formula:** Here, $a = -1$, $b = 3$, and $f(x) = g(x) = 1 - 6x^2$. 4. **Set up the integral:** $$\frac{1}{3 - (-1)} \int_{-1}^3 (1 - 6x^2) \, dx = \frac{1}{4} \int_{-1}^3 (1 - 6x^2) \, dx$$ 5. **Integrate the function:** $$\int (1 - 6x^2) \, dx = \int 1 \, dx - 6 \int x^2 \, dx = x - 6 \cdot \frac{x^3}{3} + C = x - 2x^3 + C$$ 6. **Evaluate the definite integral:** $$\int_{-1}^3 (1 - 6x^2) \, dx = \left[ x - 2x^3 \right]_{-1}^3 = (3 - 2 \cdot 3^3) - (-1 - 2 \cdot (-1)^3)$$ Calculate each term: $$3 - 2 \cdot 27 = 3 - 54 = -51$$ $$-1 - 2 \cdot (-1) = -1 + 2 = 1$$ So, $$-51 - 1 = -52$$ 7. **Calculate the average value:** $$\frac{1}{4} \times (-52) = -13$$ **Final answer:** The average value of $g(x)$ on $[-1,3]$ is **$-13$**.