1. **State the problem:** We want to use the Taylor series expansion about $x=0$ to determine whether the function $f(x) = \sin^3(x^3)$ has a local maximum, local minimum, or neither at the origin. 2. **Recall the Taylor series:** The Taylor series of a function $f(x)$ about $x=0$ is given by $$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 + \cdots$$ A critical point occurs where $f'(0) = 0$. To classify the critical point, we look at the first nonzero derivative of higher order. 3. **Find $f(0)$:** $$f(0) = \sin^3(0^3) = \sin^3(0) = 0$$ 4. **Find the first derivative $f'(x)$:** Using the chain rule, $$f(x) = (\sin(x^3))^3$$ Let $g(x) = \sin(x^3)$, then $f(x) = g(x)^3$. $$f'(x) = 3g(x)^2 \cdot g'(x)$$ Calculate $g'(x)$: $$g'(x) = \cos(x^3) \cdot 3x^2$$ So, $$f'(x) = 3 \sin^2(x^3) \cdot \cos(x^3) \cdot 3x^2 = 9x^2 \sin^2(x^3) \cos(x^3)$$ 5. **Evaluate $f'(0)$:** $$f'(0) = 9 \cdot 0^2 \cdot \sin^2(0) \cdot \cos(0) = 0$$ So, $x=0$ is a critical point. 6. **Use Taylor expansion of $\sin(x)$ near 0:** $$\sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots$$ Therefore, $$\sin(x^3) = x^3 - \frac{x^9}{6} + \cdots$$ 7. **Compute $f(x) = \sin^3(x^3)$ expansion:** $$f(x) = (\sin(x^3))^3 = \left(x^3 - \frac{x^9}{6} + \cdots \right)^3$$ The leading term is: $$x^3^3 = x^9$$ Higher order terms are of order $x^{15}$ and beyond. So, $$f(x) = x^9 + \text{higher order terms}$$ 8. **Analyze the sign near 0:** Since the leading term is $x^9$, which is an odd power, the function behaves like $x^9$ near zero. - For $x>0$, $x^9 > 0$ - For $x<0$, $x^9 < 0$ This means $f(x)$ changes sign around 0. 9. **Conclusion:** Because the function changes sign around $x=0$ and the first nonzero derivative is of odd order, $x=0$ is a critical point but neither a local maximum nor a local minimum. **Final answer:** $x=0$ is a critical point of $f$, but it is neither a local maximum nor a local minimum.