1. **State the problem:** Find the equation of the normal line to the function $f(x) = x^3 - 3x + 1$ at its local maximum point. 2. **Find the local maximum point:** - First, find the derivative $f'(x)$ to locate critical points. $$f'(x) = 3x^2 - 3$$ - Set $f'(x) = 0$ to find critical points: $$3x^2 - 3 = 0$$ $$3(x^2 - 1) = 0$$ $$\cancel{3}(x^2 - 1) = 0$$ $$x^2 - 1 = 0$$ $$x^2 = 1$$ $$x = \pm 1$$ 3. **Determine which critical point is a local maximum:** - Find the second derivative: $$f''(x) = 6x$$ - Evaluate $f''(x)$ at $x = -1$: $$f''(-1) = 6(-1) = -6 < 0$$ - Since $f''(-1) < 0$, $x = -1$ is a local maximum. 4. **Find the coordinates of the local maximum point:** $$f(-1) = (-1)^3 - 3(-1) + 1 = -1 + 3 + 1 = 3$$ So the local maximum point is $(-1, 3)$. 5. **Find the slope of the tangent line at $x = -1$:** $$f'(-1) = 3(-1)^2 - 3 = 3(1) - 3 = 0$$ 6. **Find the slope of the normal line:** - The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the tangent slope. - Since the tangent slope is $0$, the normal slope is undefined, meaning the normal line is vertical. 7. **Write the equation of the normal line:** - A vertical line through $x = -1$ is: $$x = -1$$ **Final answer:** The equation of the normal line at the local maximum point is $x = -1$.