Critical Points 93Fd59
1. **Stating the problem:**
Given the function $$Y = \frac{6}{5} x^5 - \frac{2}{7} x^6$$, we want to analyze it, find its critical points, and understand its behavior.
2. **Formula and rules:**
To find critical points, we use the derivative $$Y' = \frac{dY}{dx}$$ and set it equal to zero. Critical points occur where $$Y' = 0$$ or where $$Y'$$ is undefined.
3. **Find the derivative:**
$$Y' = \frac{d}{dx} \left( \frac{6}{5} x^5 - \frac{2}{7} x^6 \right) = \frac{6}{5} \cdot 5 x^{4} - \frac{2}{7} \cdot 6 x^{5} = 6 x^{4} - \frac{12}{7} x^{5}$$
4. **Set the derivative equal to zero:**
$$6 x^{4} - \frac{12}{7} x^{5} = 0$$
5. **Factor out common terms:**
$$x^{4} \left(6 - \frac{12}{7} x \right) = 0$$
6. **Solve for $$x$$:**
- Either $$x^{4} = 0 \Rightarrow x = 0$$
- Or $$6 - \frac{12}{7} x = 0 \Rightarrow 6 = \frac{12}{7} x \Rightarrow x = \frac{6 \times 7}{12} = \frac{42}{12} = \frac{7}{2} = 3.5$$
7. **Interpretation:**
The critical points are at $$x = 0$$ and $$x = 3.5$$.
8. **Summary:**
The function $$Y = \frac{6}{5} x^5 - \frac{2}{7} x^6$$ has critical points at $$x=0$$ and $$x=3.5$$ where the slope of the curve is zero, indicating potential maxima, minima, or points of inflection.
**Final answer:** Critical points at $$x=0$$ and $$x=3.5$$.