Discussion
Function \(f(x)\) by Maclaurin’s series (as an infinite series) can be expressed as
Answer: B
The Maclaurin series is \(f(x) = f(0) + xf'(0) + \frac{x^{2}}{2!}f''(0) + \frac{x^{3}}{3!}f'''(0) + \cdots\).
Maclaurin’s series is the Taylor series expanded about \(x = 0\). The general Taylor series about a point \(a\) is
\[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!}(x - a)^{k} \]
and with \(a = 0\) every term has the derivative evaluated at 0 and a power of \(x\), all with plus signs.
The options evaluating derivatives at 1 are not Maclaurin series, and a Taylor series about 1 would need powers of \((x - 1)\), not \(x\). The alternating-sign version at 0 is the series for \(f(-x)\), not \(f(x)\).
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