1. **State the problem:** Find all relative extrema (relative maxima and minima) of the function $$f(x) = x^4 - 32x + 3$$. 2. **Formula and rules:** To find relative extrema, we first find the critical points by setting the first derivative $$f'(x)$$ equal to zero. Then, we use the second derivative test to classify each critical point. 3. **Find the first derivative:** $$f'(x) = \frac{d}{dx}(x^4 - 32x + 3) = 4x^3 - 32$$ 4. **Set the first derivative equal to zero to find critical points:** $$4x^3 - 32 = 0$$ $$4x^3 = 32$$ $$\cancel{4}x^3 = \cancel{4}8$$ $$x^3 = 8$$ 5. **Solve for x:** $$x = \sqrt[3]{8} = 2$$ 6. **Find the second derivative:** $$f''(x) = \frac{d}{dx}(4x^3 - 32) = 12x^2$$ 7. **Evaluate the second derivative at the critical point:** $$f''(2) = 12(2)^2 = 12 \times 4 = 48$$ Since $$f''(2) > 0$$, the function is concave up at $$x=2$$, so this point is a relative minimum. 8. **Find the y-coordinate of the relative minimum:** $$f(2) = (2)^4 - 32(2) + 3 = 16 - 64 + 3 = -45$$ 9. **Check for other critical points:** Since the cubic equation $$4x^3 - 32 = 0$$ has only one real root, there is only one critical point. 10. **Conclusion:** - Relative maximum: DNE (does not exist) - Relative minimum: $$(2, -45)$$