Question

Finding Taylor and Maclaurin SeriesIn Exercises 25–34, find the Taylor series generated by f at x = a.f(x) = x³ − 2x + 4, a = 2

Accepted Answer

Identify the function and the point about which the Taylor series is generated: here, the function is $$f(x) = x$$^{3}$$ - 2x + 4$$ and the center is $$a = 2. $$Recall the formula for the Taylor series of a function $$f$$ about $$x = a$$:

$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^{n}$$

where $$f^{(n)}(a$$)$$ is the n$-th derivative of f$ evaluated at x = a$. Compute the derivatives of f(x$$)$$ up to the order needed (usually until the derivatives become zero or a pattern emerges):

- f(x$$) = $$x^{3} - 2x + 4$
- f$$'(x) = $$3x^{2} - 2$
- f$$''(x) = 6x$$
- f$$^{(3)}$$(x) = 6$$
- $$f^{(4)}(x) = 0$ (and all higher derivatives are zero). Evaluate each derivative at x = 2$:

- f(2$$) = $$2^{3} - 2(2) + 4$
- f$$'(2) = $$3(2)^{2} - 2$
- f$$''(2) = 6(2)$$
- f$$^{(3)}$$(2) = 6$$
- $$f^{(4)}(2) = 0$. Write the Taylor series expansion using the formula and the evaluated derivatives:

f(x) = f(2$$) + \frac{f'(2)}{1!}(x - 2) + \frac{f''(2)}{2!}(x - 2)^{2} + \frac{f^{(3)}(2)}{3!}(x - 2)^{3} + \cdots$$

Since higher derivatives are zero, the series will terminate here.

Finding Taylor and Maclaurin Series
In Exercises 25–34, find the Taylor series generated by f at x = a.
f(x) = x³ − 2x + 4,a = 2

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Key Concepts

Taylor Series

Maclaurin Series

Derivatives and Their Role in Series Expansion

Finding Taylor and Maclaurin SeriesIn Exercises 25–34, find the Taylor series generated by f at x = a.f(x) = x³ − 2x + 4,a = 2