1. The problem is to find the extrema of the function $$y = (x - 2)^4 - 1$$ using higher order derivatives. 2. To find extrema, we first find the first derivative $$y'$$ and set it equal to zero to find critical points. 3. Then, we use the second derivative $$y''$$ to determine the nature of each critical point: if $$y''(x) > 0$$, it's a local minimum; if $$y''(x) < 0$$, it's a local maximum; if $$y''(x) = 0$$, higher order derivatives are used. 4. Calculate the first derivative: $$y' = \frac{d}{dx} \left((x - 2)^4 - 1\right) = 4(x - 2)^3$$ 5. Set the first derivative equal to zero to find critical points: $$4(x - 2)^3 = 0$$ $$\Rightarrow (x - 2)^3 = 0$$ $$\Rightarrow x = 2$$ 6. Calculate the second derivative: $$y'' = \frac{d}{dx} (4(x - 2)^3) = 12(x - 2)^2$$ 7. Evaluate the second derivative at the critical point: $$y''(2) = 12(2 - 2)^2 = 12 \times 0 = 0$$ Since $$y''(2) = 0$$, the second derivative test is inconclusive. 8. Calculate the third derivative: $$y''' = \frac{d}{dx} (12(x - 2)^2) = 24(x - 2)$$ 9. Evaluate the third derivative at $$x=2$$: $$y'''(2) = 24(2 - 2) = 0$$ 10. Calculate the fourth derivative: $$y^{(4)} = \frac{d}{dx} (24(x - 2)) = 24$$ 11. Since $$y^{(4)}(2) = 24 > 0$$ and the first non-zero derivative after the second derivative is of even order and positive, the point $$x=2$$ is a local minimum. 12. Find the minimum value: $$y(2) = (2 - 2)^4 - 1 = 0 - 1 = -1$$ **Final answer:** The function has a local minimum at $$x=2$$ with value $$y = -1$$.