AP Calculus BC : Polynomial Approximations and Series

Study concepts, example questions & explanations for AP Calculus BC

varsity tutors app store varsity tutors android store

Example Questions

Example Question #11 : Maclaurin Series

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

Possible Answers:

\(\displaystyle 480.00\)

\(\displaystyle 160.00\)

\(\displaystyle -8.89\)

\(\displaystyle 80.00\)

Correct answer:

\(\displaystyle 80.00\)

Explanation:

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

Example Question #12 : Maclaurin Series

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

Possible Answers:

\(\displaystyle 15.43\)

\(\displaystyle -370.44\)

\(\displaystyle -92.61\)

\(\displaystyle -10.29\)

Correct answer:

\(\displaystyle -92.61\)

Explanation:

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

Example Question #13 : Maclaurin Series

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

Possible Answers:

\(\displaystyle 8.14\)

\(\displaystyle 73.24\)

\(\displaystyle -366.20\)

\(\displaystyle -24.41\)

Correct answer:

\(\displaystyle 73.24\)

Explanation:

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

Example Question #14 : Maclaurin Series

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

Possible Answers:

\(\displaystyle 1.60\)

\(\displaystyle -14.44\)

\(\displaystyle 2.06\)

\(\displaystyle -3.61\)

Correct answer:

\(\displaystyle -14.44\)

Explanation:

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

Example Question #41 : Ap Calculus Bc

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

Possible Answers:

\(\displaystyle 4.50\)

\(\displaystyle 2.00\)

\(\displaystyle -18.00\)

\(\displaystyle -108.00\)

Correct answer:

\(\displaystyle -18.00\)

Explanation:

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

Example Question #41 : Ap Calculus Bc

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

Possible Answers:

\(\displaystyle -2940\)

\(\displaystyle 1079\)

\(\displaystyle 647\)

\(\displaystyle 348\)

Correct answer:

\(\displaystyle 647\)

Explanation:

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

Example Question #42 : Ap Calculus Bc

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

Possible Answers:

\(\displaystyle -\frac{32}{3}\)

\(\displaystyle -22\)

\(\displaystyle -5\)

\(\displaystyle -126\)

Correct answer:

\(\displaystyle -22\)

Explanation:

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

Example Question #43 : Ap Calculus Bc

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

Possible Answers:

\(\displaystyle -\frac{99}{2}\)

\(\displaystyle 0\)

\(\displaystyle -111\)

\(\displaystyle -66\)

Correct answer:

\(\displaystyle -66\)

Explanation:

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

Example Question #44 : Ap Calculus Bc

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

Possible Answers:

\(\displaystyle -\frac{36619}{24}\)

\(\displaystyle -460\)

\(\displaystyle -275\)

\(\displaystyle -3364\)

Correct answer:

\(\displaystyle -460\)

Explanation:

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

Example Question #45 : Ap Calculus Bc

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

Possible Answers:

\(\displaystyle 30\)

\(\displaystyle 14\)

\(\displaystyle \frac{50}{3}\)

\(\displaystyle 46\)

Correct answer:

\(\displaystyle 30\)

Explanation:

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

Learning Tools by Varsity Tutors