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AP Calculus BC : Maclaurin Series

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Example Questions

Example Question #21 : Maclaurin 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:

4.50

-108.00

-18.00

2.00

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 #22 : Maclaurin 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

348

647

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 #23 : Maclaurin 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 #24 : Maclaurin 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 #25 : Maclaurin Series

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

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 #26 : Maclaurin Series

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

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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Example Question #27 : Maclaurin Series

$\begin{align*}&\text{Approximate }f(-8) \\&\text{Where }f(0)=-20;f'(0)=-3;f''(0)=6\\&\text{With a Maclaurin series expansion.}\end{align*}$

Possible Answers:

160

388

196

Correct answer:

196

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function values }f(0)=-20;f'(0)=-3;f''(0)=6\\&\text{ at }x=-8\\&\text{This expansion becomes:}\\&f(-8)\approx \frac{-20}{0!}(-8)^{0}+\frac{-3}{1!}(-8)^{1}+\frac{6}{2!}(-8)^{2}\\&f(-8)\approx196\end{align*}$

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Example Question #28 : Maclaurin Series

$\begin{align*}&\text{Approximate }f(-4) \\&\text{Where }f(0)=-12;f'(0)=-10;f''(0)=16;f'''(0)=8\\&\text{By using a Maclaurin expansion.}\end{align*}$

Possible Answers:

$\frac{212}{3}$

$\frac{152}{3}$

$-\frac{352}{3}$

Correct answer:

$\frac{212}{3}$

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function values }f(0)=-12;f'(0)=-10;f''(0)=16;f'''(0)=8\\&\text{ at }x=-4\\&\text{This expansion becomes:}\\&f(-4)\approx \frac{-12}{0!}(-4)^{0}+\frac{-10}{1!}(-4)^{1}+\frac{16}{2!}(-4)^{2}+\frac{8}{3!}(-4)^{3}\\&f(-4)\approx\frac{212}{3}\end{align*}$

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Example Question #29 : Maclaurin Series

$\begin{align*}&\text{Approximate }f(-7) \\&\text{Where }f(0)=-17;f'(0)=17;f''(0)=8\\&\text{By using a Maclaurin expansion.}\end{align*}$

Possible Answers:

$\frac{469}{6}$

256

Correct answer:

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function values }f(0)=-17;f'(0)=17;f''(0)=8\\&\text{ at }x=-7\\&\text{This expansion becomes:}\\&f(-7)\approx \frac{-17}{0!}(-7)^{0}+\frac{17}{1!}(-7)^{1}+\frac{8}{2!}(-7)^{2}\\&f(-7)\approx60\end{align*}$

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Example Question #30 : Maclaurin Series

$\begin{align*}&\text{For the function values: }\\&f(0)=5;f'(0)=20;f''(0)=-14\\&\text{Approximate }f(-4)\text{ using a Maclaurin expansion.}\end{align*}$

Possible Answers:

$\frac{868}{3}$

Correct answer:

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function values }f(0)=5;f'(0)=20;f''(0)=-14\\&\text{ at }x=-4\\&\text{This expansion becomes:}\\&f(-4)\approx \frac{5}{0!}(-4)^{0}+\frac{20}{1!}(-4)^{1}+\frac{-14}{2!}(-4)^{2}\\&f(-4)\approx-187\end{align*}$

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AP Calculus BC : Maclaurin Series

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #21 : Maclaurin Series

Example Question #22 : Maclaurin Series

Example Question #23 : Maclaurin Series

Example Question #24 : Maclaurin Series

Example Question #25 : Maclaurin Series

Example Question #26 : Maclaurin Series

Example Question #27 : Maclaurin Series

Example Question #28 : Maclaurin Series

Example Question #29 : Maclaurin Series

Example Question #30 : Maclaurin Series

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