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AP Calculus BC : Polynomial Approximations and Series

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Example Questions

Example Question #11 : Maclaurin Series

$\begin{align*}&\text{Approximate }f(3) \\&\text{Where }f(x)=-cos(5x)e^{(-3x)}\\&\text{With a Taylor series centered at zero, using }3\text{ terms}\end{align*}$

Possible Answers:

480.00

-8.89

80.00

160.00

Correct answer:

80.00

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function }-cos(5x)e^{(-3x)}\text{ at }x=3\\&\text{This expansion becomes:}\\&f(3)\approx (-cos(5(0))e^{(-3(0))})(3)^{0}+(3cos(5(0))e^{(-3(0))} + 5sin(5(0))e^{(-3(0))})(3)^{1}+\\&\frac{16cos(5(0))e^{(-3(0))} - 30sin(5(0))e^{(-3(0))}}{2}(3)^{2}\\&f(3)\approx80.00\end{align*}$

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Example Question #12 : Maclaurin Series

$\begin{align*}&\text{Approximate }f(2.1) \\&\text{Where }f(x)=-sin(3x)sin(7x)\\&\text{By using a Maclaurin expansion to }3\text{ terms}\end{align*}$

Possible Answers:

15.43

-10.29

-370.44

-92.61

Correct answer:

-92.61

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function }-sin(3x)sin(7x)\text{ at }x=2.1\\&\text{This expansion becomes:}\\&f(2.1)\approx (-sin(3(0))sin(7(0)))(2.1)^{0}+(- 3cos(3(0))sin(7(0)) - 7cos(7(0))sin(3(0)))(2.1)^{1}+\\&\frac{58sin(3(0))sin(7(0)) - 42cos(3(0))cos(7(0))}{2}(2.1)^{2}\\&f(2.1)\approx-92.61\end{align*}$

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Example Question #41 : Taylor Series

$\begin{align*}&\text{Approximate }f(1.6) \\&\text{Where }f(x)=-cos(3x)cos(7x)\\&\text{With a Taylor series centered at zero, using }3\text{ terms}\end{align*}$

Possible Answers:

-24.41

8.14

-366.20

73.24

Correct answer:

73.24

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function }-cos(3x)cos(7x)\text{ at }x=1.6\\&\text{This expansion becomes:}\\&f(1.6)\approx (-cos(3(0))cos(7(0)))(1.6)^{0}+(7cos(3(0))sin(7(0)) + 3cos(7(0))sin(3(0)))(1.6)^{1}+\\&\frac{58cos(3(0))cos(7(0)) - 42sin(3(0))sin(7(0))}{2}(1.6)^{2}\\&f(1.6)\approx73.24\end{align*}$

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Example Question #14 : Maclaurin Series

$\begin{align*}&\text{For the function: }f(x)=-sin(2x)^{2}\\&\text{Approximate }f(1.9)\\&\text{Using a Maclaurin expansion to }3\text{ terms}\end{align*}$

Possible Answers:

1.60

2.06

-3.61

-14.44

Correct answer:

-14.44

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function }-sin(2x)^{2}\text{ at }x=1.9\\&\text{This expansion becomes:}\\&f(1.9)\approx (-sin(2(0))^{2})(1.9)^{0}+(-4cos(2(0))sin(2(0)))(1.9)^{1}+\\&\frac{8sin(2(0))^{2} - 8cos(2(0))^{2}}{2}(1.9)^{2}\\&f(1.9)\approx-14.44\end{align*}$

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Example Question #41 : Taylor Series

$\begin{align*}&\text{Approximate }f(3.6) \\&\text{Where }f(x)=-cos(5x)sin(5x)\\&\text{With a Taylor series centered at zero, using }3\text{ terms}\end{align*}$

Possible Answers:

2.00

4.50

-108.00

-18.00

Correct answer:

-18.00

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function }-cos(5x)sin(5x)\text{ at }x=3.6\\&\text{This expansion becomes:}\\&f(3.6)\approx (-cos(5(0))sin(5(0)))(3.6)^{0}+(5sin(5(0))^{2} - 5cos(5(0))^{2})(3.6)^{1}+\\&\frac{100cos(5(0))sin(5(0))}{2}(3.6)^{2}\\&f(3.6)\approx-18.00\end{align*}$

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Example Question #41 : Polynomial Approximations And Series

$\begin{align*}&\text{For the function values: }\\&f(0)=-13;f'(0)=-19;f''(0)=6\\&\text{Approximate }f(-12)\\&\text{Using a Maclaurin expansion.}\end{align*}$

Possible Answers:

1079

647

348

Correct answer:

647

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function values }f(0)=-13;f'(0)=-19;f''(0)=6\\&\text{ at }x=-12\\&\text{This expansion becomes:}\\&f(-12)\approx \frac{-13}{0!}(-12)^{0}+\frac{-19}{1!}(-12)^{1}+\frac{6}{2!}(-12)^{2}\\&f(-12)\approx647\end{align*}$

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Example Question #43 : Taylor Series

$\begin{align*}&\text{Approximate }f(2) \\&\text{Where }f(0)=0;f'(0)=1;f''(0)=-2;f'''(0)=-15\\&\text{By using a Maclaurin expansion.}\end{align*}$

Possible Answers:

$-\frac{32}{3}$

Correct answer:

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function values }f(0)=0;f'(0)=1;f''(0)=-2;f'''(0)=-15\\&\text{ at }x=2\\&\text{This expansion becomes:}\\&f(2)\approx \frac{0}{0!}(2)^{0}+\frac{1}{1!}(2)^{1}+\frac{-2}{2!}(2)^{2}+\frac{-15}{3!}(2)^{3}\\&f(2)\approx-22\end{align*}$

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Example Question #41 : Taylor Series

$\begin{align*}&\text{Approximate }f(3) \\&\text{Where }f(0)=18;f'(0)=-13;f''(0)=-10\\&\text{With a Maclaurin series expansion.}\end{align*}$

Possible Answers:

$-\frac{99}{2}$

Correct answer:

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function values }f(0)=18;f'(0)=-13;f''(0)=-10\\&\text{ at }x=3\\&\text{This expansion becomes:}\\&f(3)\approx \frac{18}{0!}(3)^{0}+\frac{-13}{1!}(3)^{1}+\frac{-10}{2!}(3)^{2}\\&f(3)\approx-66\end{align*}$

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Example Question #43 : Ap Calculus Bc

$\begin{align*}&\text{Approximate }f(11) \\&\text{Where }f(0)=-9;f'(0)=-19;f''(0)=7;f'''(0)=-3\\&\text{By using a Maclaurin expansion.}\end{align*}$

Possible Answers:

$-\frac{36619}{24}$

Correct answer:

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function values }f(0)=-9;f'(0)=-19;f''(0)=7;f'''(0)=-3\\&\text{ at }x=11\\&\text{This expansion becomes:}\\&f(11)\approx \frac{-9}{0!}(11)^{0}+\frac{-19}{1!}(11)^{1}+\frac{7}{2!}(11)^{2}+\frac{-3}{3!}(11)^{3}\\&f(11)\approx-460\end{align*}$

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Example Question #41 : Ap Calculus Bc

$\begin{align*}&\text{For the function values: }\\&f(0)=-8;f'(0)=11;f''(0)=8\\&\text{Approximate }f(2)\text{ using a Maclaurin expansion.}\end{align*}$

Possible Answers:

$\frac{50}{3}$

Correct answer:

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function values }f(0)=-8;f'(0)=11;f''(0)=8\\&\text{ at }x=2\\&\text{This expansion becomes:}\\&f(2)\approx \frac{-8}{0!}(2)^{0}+\frac{11}{1!}(2)^{1}+\frac{8}{2!}(2)^{2}\\&f(2)\approx30\end{align*}$

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AP Calculus BC : Polynomial Approximations and Series

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #11 : Maclaurin Series

Example Question #12 : Maclaurin Series

Example Question #41 : Taylor Series

Example Question #14 : Maclaurin Series

Example Question #41 : Taylor Series

Example Question #41 : Polynomial Approximations And Series

Example Question #43 : Taylor Series

Example Question #41 : Taylor Series

Example Question #43 : Ap Calculus Bc

Example Question #41 : Ap Calculus Bc

All AP Calculus BC Resources

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