4. Let f:R->R be a function with Taylor series converging to f(x) for all real numbers x. If f(0) = 2, f’(0) = 2, and f(n)(0) = 3 for n >= 2, then f(x) =
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A. 3ex + 2x - 1 |
B. e3x + 2x + 1 |
C. e3x - x + 1 |
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D. 3ex - x - 1 |
E. 3ex + 5x + 5 |
Solution: The answer is D
Since f(x) has a Taylor series converging to f(x) for all real numbers x, consider the Taylor series of f(x) centered at 0, such that,
Substituting the given values, f(0) = 2, f’(0) = 2, and f(n)(0) = 3 for n >= 2, into the above equation yields:
Since,
Thus,
