1. **State the problem:** Find the minima and maxima of the function $$y=(x-4)^4(x+3)^3$$. 2. **Formula and rules:** To find minima and maxima, we first find the critical points by setting the first derivative $$y'$$ equal to zero. Then, we use the second derivative test to classify each critical point. 3. **Find the first derivative:** $$y = (x-4)^4 (x+3)^3$$ Using the product rule: $$y' = 4(x-4)^3 (x+3)^3 + (x-4)^4 \cdot 3(x+3)^2$$ 4. **Factor the first derivative:** $$y' = (x-4)^3 (x+3)^2 [4(x+3) + 3(x-4)]$$ Simplify inside the bracket: $$4x + 12 + 3x - 12 = 7x$$ So, $$y' = 7x (x-4)^3 (x+3)^2$$ 5. **Find critical points by setting $$y' = 0$$:** $$7x (x-4)^3 (x+3)^2 = 0$$ Critical points are where any factor is zero: $$x=0, x=4, x=-3$$ 6. **Find the second derivative $$y''$$ to classify critical points:** We differentiate $$y'$$: $$y' = 7x (x-4)^3 (x+3)^2$$ Using product rule and chain rule (detailed steps omitted for brevity), the sign of $$y''$$ at each critical point determines minima or maxima. 7. **Evaluate $$y''$$ at critical points:** - At $$x=0$$, $$y''(0) < 0$$ (negative), so $$x=0$$ is a local maximum. - At $$x=4$$, $$y''(4) = 0$$ and higher derivatives show it is a point of inflection, not a max or min. - At $$x=-3$$, $$y''(-3) = 0$$ and the function behavior shows a point of inflection. 8. **Conclusion:** - Local maximum at $$x=0$$. - No local minima. - Points $$x=4$$ and $$x=-3$$ are points of inflection. Final answer: $$\boxed{\text{Local maximum at } x=0, \text{ no local minima}}$$