1. **Problem Statement:** Determine the Taylor series for the function $f(x) = x^3$ at $a=2$. 2. **Taylor Series Formula:** The Taylor series of a function $f(x)$ at $x=a$ is given by: $$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x - a)^n$$ where $f^{(n)}(a)$ is the $n$th derivative of $f$ evaluated at $a$. 3. **Calculate derivatives of $f(x) = x^3$:** - $f(x) = x^3$ - $f'(x) = 3x^2$ - $f''(x) = 6x$ - $f^{(3)}(x) = 6$ - $f^{(n)}(x) = 0$ for $n \geq 4$ 4. **Evaluate derivatives at $a=2$:** - $f(2) = 2^3 = 8$ - $f'(2) = 3 \times 2^2 = 12$ - $f''(2) = 6 \times 2 = 12$ - $f^{(3)}(2) = 6$ 5. **Write the Taylor series:** $$ \begin{aligned} f(x) &= f(2) + f'(2)(x-2) + \frac{f''(2)}{2!}(x-2)^2 + \frac{f^{(3)}(2)}{3!}(x-2)^3 \\ &= 8 + 12(x-2) + \frac{12}{2}(x-2)^2 + \frac{6}{6}(x-2)^3 \\ &= 8 + 12(x-2) + 6(x-2)^2 + (x-2)^3 \end{aligned} $$ 6. **Explanation:** The Taylor series expresses $f(x)$ as a polynomial centered at $a=2$ using derivatives at that point. Since $f(x) = x^3$ is a polynomial of degree 3, the Taylor series terminates at the third derivative. **Final answer:** $$f(x) = 8 + 12(x-2) + 6(x-2)^2 + (x-2)^3$$