1. **State the problem:** We want to find the Taylor series expansion of the function $$f(x) = e^{\sin(x^4)} \cos(x^2)$$ about $$x=0$$ and determine the nature of the critical point at $$x=0$$. 2. **Recall Taylor series expansions:** - $$\sin(x) = x - \frac{x^3}{6} + O(x^5)$$ - $$e^x = 1 + x + \frac{x^2}{2} + O(x^3)$$ - $$\cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$$ 3. **Expand inner functions:** - $$\sin(x^4) = x^4 - \frac{x^{12}}{6} + O(x^{20})$$ 4. **Expand $$e^{\sin(x^4)}$$:** Using $$e^y = 1 + y + \frac{y^2}{2} + O(y^3)$$ with $$y = \sin(x^4)$$, $$e^{\sin(x^4)} = 1 + x^4 + \frac{x^8}{2} + O(x^{12})$$ 5. **Expand $$\cos(x^2)$$:** $$\cos(x^2) = 1 - \frac{x^4}{2} + \frac{x^8}{24} + O(x^{12})$$ 6. **Multiply expansions:** $$f(x) = e^{\sin(x^4)} \cos(x^2) = \left(1 + x^4 + \frac{x^8}{2} + O(x^{12})\right) \left(1 - \frac{x^4}{2} + \frac{x^8}{24} + O(x^{12})\right)$$ Multiply terms up to order $$x^8$$: $$= 1 + x^4 - \frac{x^4}{2} + \frac{x^8}{2} + \frac{x^8}{24} - \frac{x^8}{2} + O(x^{12})$$ Simplify: $$= 1 + \frac{x^4}{2} + \left(\frac{1}{2} + \frac{1}{24} - \frac{1}{2}\right) x^8 + O(x^{12}) = 1 + \frac{x^4}{2} + \frac{x^8}{24} + O(x^{12})$$ 7. **Interpretation:** - The Taylor series about $$x=0$$ is $$1 + \frac{x^4}{2} + O(x^5)$$. - The first derivative at $$x=0$$ is zero (no $$x$$ or $$x^3$$ terms). - The second derivative at $$x=0$$ is zero (no $$x^2$$ term). - The first nonzero term after the constant is $$\frac{x^4}{2}$$, which is positive. 8. **Conclusion:** Since the first nonzero derivative after the constant term is the fourth derivative and it is positive, $$x=0$$ is a local minimum. **Final answer:** The correct statement is: "Its Taylor series expansion about $$x=0$$ is $$1 + \frac{x^4}{2} + O(x^5)$$. Hence it has a local minimum at $$x=0$$."