1. **State the problem:** We have the function $x = \frac{1}{6}k^3 - 4k$. We need to find: a) A value of $k$ such that $\frac{dx}{dk} = 46$. b) The corresponding value of $x$ at that $k$. 2. **Find the derivative $\frac{dx}{dk}$:** Using the power rule, $\frac{d}{dk} \left( \frac{1}{6}k^3 \right) = \frac{1}{6} \times 3k^2 = \frac{1}{2}k^2$. Also, $\frac{d}{dk}(-4k) = -4$. So, $$\frac{dx}{dk} = \frac{1}{2}k^2 - 4$$ 3. **Set the derivative equal to 46 and solve for $k$:** $$\frac{1}{2}k^2 - 4 = 46$$ Add 4 to both sides: $$\frac{1}{2}k^2 = 50$$ Multiply both sides by 2: $$\cancel{2} \times \frac{1}{2}k^2 = 50 \times \cancel{2}$$ $$k^2 = 100$$ 4. **Solve for $k$:** $$k = \pm \sqrt{100} = \pm 10$$ 5. **Find $x$ when $k=10$:** $$x = \frac{1}{6} (10)^3 - 4(10) = \frac{1}{6} \times 1000 - 40 = 166.666\ldots - 40 = 126.666\ldots$$ Rounded to 1 decimal place: $$x = 126.7$$ 6. **Find $x$ when $k=-10$:** $$x = \frac{1}{6} (-10)^3 - 4(-10) = \frac{1}{6} \times (-1000) + 40 = -166.666\ldots + 40 = -126.666\ldots$$ Rounded to 1 decimal place: $$x = -126.7$$ **Final answers:** - $k = 10$ or $k = -10$ when $\frac{dx}{dk} = 46$. - Corresponding $x$ values are $126.7$ and $-126.7$ respectively.