1. **Problem:** Find the maximum and minimum values of the function $y = x^3 - 3x + 2$. 2. **Formula and rules:** To find maxima and minima, we use the first derivative test. Find $y'$ and set it equal to zero to find critical points. 3. **Calculate the derivative:** $$y' = \frac{d}{dx}(x^3 - 3x + 2) = 3x^2 - 3$$ 4. **Find critical points:** Set $y' = 0$: $$3x^2 - 3 = 0 \implies x^2 = 1 \implies x = \pm 1$$ 5. **Second derivative test:** Calculate $y''$: $$y'' = \frac{d}{dx}(3x^2 - 3) = 6x$$ Evaluate at critical points: - At $x = 1$: $y''(1) = 6(1) = 6 > 0$, so $x=1$ is a local minimum. - At $x = -1$: $y''(-1) = 6(-1) = -6 < 0$, so $x=-1$ is a local maximum. 6. **Find function values at critical points:** - $y(1) = 1^3 - 3(1) + 2 = 1 - 3 + 2 = 0$ - $y(-1) = (-1)^3 - 3(-1) + 2 = -1 + 3 + 2 = 4$ 7. **Conclusion:** - Maximum value is $4$ at $x = -1$. - Minimum value is $0$ at $x = 1$.