AP Calculus BC 10.14: Series Interpretation

Practice AP Calculus BC questions on recognizing geometric series and using known foundational series to identify represented functions.

[Maximum number: 1]

The Taylor series for a function f about x=4 is given by

$\sum_{n=1}^{\infty} \frac{(x-4)^{n+1}}{(n+1) 3^{n}}=\frac{(x-4)^{2}}{2 \cdot 3}+\frac{(x-4)^{3}}{3 \cdot 3^{2}}+\frac{(x-4)^{4}}{4 \cdot 3^{3}}+\cdots+\frac{(x-4)^{n+1}}{(n+1) 3^{n}}+\cdots$ and converges to f(x) on

its interval of convergence.

The Taylor series for $f^{\prime}$ described in part B is a geometric series. For all x in the interval of

convergence of the Taylor series for $f^{\prime}$ , show that $f^{\prime}(x)=\frac{x-4}{7-x}$ .

C The Taylor series for $f^{\prime}$ described in part B is a geometric series. For all x in the interval of convergence

of the Taylor series for $f^{\prime}$ , show that $f^{\prime}(x)=\frac{x-4}{7-x}$ .

The Taylor series for $f^{\prime}$ is a geometric series with first term $\frac{x-4}{3}$ and common ratio $\frac{x-4}{3}$ . $f^{\prime}(x)=\frac{\frac{x-4}{3}}{1-\frac{x-4}{3}}=\frac{x-4}{3-(x-4)}=\frac{x-4}{7-x}$	Verification	Point 8 (P8)
Scoring Notes for Part C

- A response of $f^{\prime}(x)=\frac{\frac{x-4}{3}}{1-\frac{x-4}{3}}$ is sufficient to earn P8.

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LIM-8.F—Interpret Taylor series and Maclaurin series question 1