1) \( S(x)=\frac{x^{3}}{3}-9 x \) on \( [-7,7] \)

**Step 1. Find the derivative.** The function is \[ S(x)=\frac{x^3}{3}-9x. \] Differentiate with respect to \( x \): \[ S'(x)=\frac{d}{dx}\left(\frac{x^3}{3}-9x\right)=x^2-9. \] **Step 2. Find the critical points.** Set the derivative equal to zero: \[ x^2-9=0. \] Solve for \( x \): \[ x^2=9 \quad \Longrightarrow \quad x=\pm 3. \] Both \( x=-3 \) and \( x=3 \) are in the interval \([-7,7]\). **Step 3. Evaluate the function at the critical points and the endpoints.** - At \( x=-3 \): \[ S(-3)=\frac{(-3)^3}{3}-9(-3)=\frac{-27}{3}+27=-9+27=18. \] - At \( x=3 \): \[ S(3)=\frac{3^3}{3}-9(3)=\frac{27}{3}-27=9-27=-18. \] - At the left endpoint \( x=-7 \): \[ S(-7)=\frac{(-7)^3}{3}-9(-7)=\frac{-343}{3}+63=\frac{-343+189}{3}=\frac{-154}{3}\approx -51.33. \] - At the right endpoint \( x=7 \): \[ S(7)=\frac{7^3}{3}-9(7)=\frac{343}{3}-63=\frac{343-189}{3}=\frac{154}{3}\approx 51.33. \] **Step 4. Determine the absolute maximum and minimum on \([-7,7]\).** The evaluated values are: - \( S(-7)=\frac{-154}{3}\approx -51.33 \) - \( S(-3)=18 \) - \( S(3)=-18 \) - \( S(7)=\frac{154}{3}\approx 51.33 \) Thus: - The **absolute minimum** on \([-7,7]\) is \[ S(-7)=\frac{-154}{3}, \] occurring at \( x=-7 \). - The **absolute maximum** on \([-7,7]\) is \[ S(7)=\frac{154}{3}, \] occurring at \( x=7 \). Additionally, note that there is a **local maximum** at \( x=-3 \) with \( S(-3)=18 \) and a **local minimum** at \( x=3 \) with \( S(3)=-18 \). **Final Answer:** - Absolute minimum: \( S(-7)=\frac{-154}{3} \) at \( x=-7 \). - Absolute maximum: \( S(7)=\frac{154}{3} \) at \( x=7 \). - Local maximum: \( S(-3)=18 \) at \( x=-3 \). - Local minimum: \( S(3)=-18 \) at \( x=3 \).