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AP Calculus BC : Taylor Polynomial Approximation

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Example Questions

Example Question #21 : Polynomial Approximations And Series

$\begin{align*}&\text{Approximate }f(1) \\&\text{Where }f(x)=- cos(5x) - sin(2x)\\&\text{With a Taylor series centered at }\frac{(5\cdot \pi )}{2}\text{, using }3\text{ terms}\end{align*}$

Possible Answers:

7089.24

-47.98

-335.85

47.98

Correct answer:

-47.98

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function }- cos(5x) - sin(2x)\text{ at }x=1\\&\text{This expansion becomes:}\\&f(x)\approx (- cos(5a) - sin(2a))+(5sin(5a) - 2cos(2a))(x-a)^{1}+\frac{25cos(5a) + 4sin(2a)}{2}(x-a)^{2}\\&x=1\\&a=\frac{(5\cdot \pi )}{2}\\&f(1)\approx-47.98\end{align*}$

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Example Question #22 : Polynomial Approximations And Series

$\begin{align*}&\text{For the function: }f(x)=-cos(4x)^{2}\\&\text{Approximate }f(3)\\&\text{Using a Taylor expansion centered at }6\cdot \pi\text{ to }2\text{ terms}\end{align*}$

Possible Answers:

-9.00

-0.17

-1.00

4019.33

Correct answer:

-1.00

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function }-cos(4x)^{2}\text{ at }x=3\\&\text{This expansion becomes:}\\&f(x)\approx (-cos(4a)^{2})+(8cos(4a)sin(4a))(x-a)^{1}\\&x=3\\&a=6\cdot \pi\\&f(3)\approx-1.00\end{align*}$

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Example Question #21 : Taylor Series

$\begin{align*}&\text{Approximate }f(5) \\&\text{Where }f(x)=sin(6x) - cos(2x)\\&\text{By using a Taylor expansion to }3\text{ terms, centered at }\pi\end{align*}$

Possible Answers:

23.97

17.06

-5.24

-213.00

Correct answer:

17.06

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function }sin(6x) - cos(2x)\text{ at }x=5\\&\text{This expansion becomes:}\\&f(x)\approx (sin(6a) - cos(2a))+(6cos(6a) + 2sin(2a))(x-a)^{1}+\frac{4cos(2a) - 36sin(6a)}{2}(x-a)^{2}\\&x=5\\&a=\pi\\&f(5)\approx17.06\end{align*}$

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Example Question #21 : Taylor Series

$\begin{align*}&\text{Approximate }f(6) \\&\text{Where }f(x)=- cos(9x) - cos(x)\\&\text{With a Taylor series centered at }-\frac{(5\cdot \pi )}{2}\text{, using }3\text{ terms}\end{align*}$

Possible Answers:

323377.21

-138.54

138.54

692.70

Correct answer:

-138.54

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function }- cos(9x) - cos(x)\text{ at }x=6\\&\text{This expansion becomes:}\\&f(x)\approx (- cos(9a) - cos(a))+(9sin(9a) + sin(a))(x-a)^{1}+\frac{81cos(9a) + cos(a)}{2}(x-a)^{2}\\&x=6\\&a=-\frac{(5\cdot \pi )}{2}\\&f(6)\approx-138.54\end{align*}$

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AP Calculus BC : Taylor Polynomial Approximation

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #21 : Polynomial Approximations And Series

Example Question #22 : Polynomial Approximations And Series

Example Question #21 : Taylor Series

Example Question #21 : Taylor Series

All AP Calculus BC Resources

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