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AP Calculus BC : AP Calculus BC

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Example Questions

Example Question #11 : Ap Calculus Bc

$\begin{align*}&\text{For the function values: }\\&f(1)=3;f'(1)=-13;f''(1)=19;f'''(1)=-10;f''''(1)=17\\&\text{Approximate }f(-3)\text{ using a Taylor expansion.}\end{align*}$

Possible Answers:

$-\frac{2852}{5}$

$\frac{533}{3}$

495

3967

Correct answer:

495

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function values }f(1)=3;f'(1)=-13;f''(1)=19;f'''(1)=-10;f''''(1)=17\\&\text{ at }x=-3\\&\text{This expansion becomes:}\\&f(-3)\approx \frac{3}{0!}(-3-1)^{0}+\frac{-13}{1!}(-3-1)^{1}+\frac{19}{2!}(-3-1)^{2}+\frac{-10}{3!}(-3-1)^{3}+\frac{17}{4!}(-3-1)^{4}\\&f(-3)\approx495\end{align*}$

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Example Question #12 : Ap Calculus Bc

$\begin{align*}&\text{Approximate }f(-5) \\&\text{Where }f(-8)=-11;f'(-8)=12;f''(-8)=20;f'''(-8)=-19\\&\text{By using a Taylor expansion.}\end{align*}$

Possible Answers:

646

$\frac{59}{2}$

$\frac{257}{2}$

$\frac{375}{8}$

Correct answer:

$\frac{59}{2}$

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function values }f(-8)=-11;f'(-8)=12;f''(-8)=20;f'''(-8)=-19\\&\text{ at }x=-5\\&\text{This expansion becomes:}\\&f(-5)\approx \frac{-11}{0!}(-5-(-8))^{0}+\frac{12}{1!}(-5-(-8))^{1}+\frac{20}{2!}(-5-(-8))^{2}+\frac{-19}{3!}(-5-(-8))^{3}\\&f(-5)\approx\frac{59}{2}\end{align*}$

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Example Question #13 : Ap Calculus Bc

$\begin{align*}&\text{Approximate }f(6) \\&\text{Where }f(x)=-sin(4x)sin(5x)\\&\text{With a Taylor series centered at }2\cdot \pi\text{, using }3\text{ terms}\end{align*}$

Possible Answers:

1.60

-3.21

3.21

-1.60

Correct answer:

-1.60

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function }-sin(4x)sin(5x)\text{ at }x=6\\&\text{This expansion becomes:}\\&f(x)\approx (-sin(4a)sin(5a))+(- 4cos(4a)sin(5a) - 5cos(5a)sin(4a))(x-a)^{1}+\frac{41sin(4a)sin(5a) - 40cos(4a)cos(5a)}{2}(x-a)^{2}\\&x=6\\&a=2\cdot \pi\\&f(6)\approx-1.60\end{align*}$

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Example Question #14 : Ap Calculus Bc

$\begin{align*}&\text{Approximate }f(1) \\&\text{Where }f(x)=- sin(6x) - sin(7x)\\&\text{With a Taylor series centered at }\frac{(15\cdot \pi )}{4}\text{, using }3\text{ terms}\end{align*}$

Possible Answers:

4050.63

-46493.53

8263.06

4157.36

Correct answer:

4157.36

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function }- sin(6x) - sin(7x)\text{ at }x=1\\&\text{This expansion becomes:}\\&f(x)\approx (- sin(6a) - sin(7a))+(- 6cos(6a) - 7cos(7a))(x-a)^{1}+\frac{36sin(6a) + 49sin(7a)}{2}(x-a)^{2}\\&x=1\\&a=\frac{(15\cdot \pi )}{4}\\&f(1)\approx4157.36\end{align*}$

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Example Question #11 : Taylor Polynomial Approximation

$\begin{align*}&\text{For the function: }f(x)=sin(9x)sin(x)\\&\text{Approximate }f(4)\\&\text{Using a Taylor expansion centered at }\frac{(15\cdot \pi )}{4}\text{ to }2\text{ terms}\end{align*}$

Possible Answers:

-38.40

-929.79

-5.63

39.40

Correct answer:

39.40

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function }sin(9x)sin(x)\text{ at }x=4\\&\text{This expansion becomes:}\\&f(x)\approx (sin(9a)sin(a))+(9cos(9a)sin(a) + sin(9a)cos(a))(x-a)^{1}\\&x=4\\&a=\frac{(15\cdot \pi )}{4}\\&f(4)\approx39.40\end{align*}$

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Example Question #12 : Taylor Polynomial Approximation

$\begin{align*}&\text{Approximate }f(10) \\&\text{Where }f(x)=\frac{cos(5x)}{sin(5x)}\\&\text{By using a Taylor expansion to }2\text{ terms, centered at }\frac{\pi }{2}\end{align*}$

Possible Answers:

-42.15

42.15

-8.43

8.43

Correct answer:

-42.15

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function }\frac{cos(5x)}{sin(5x)}\text{ at }x=10\\&\text{This expansion becomes:}\\&f(x)\approx (\frac{cos(5a)}{sin(5a)})+(-\frac{ (5cos(5a)^{2})}{sin(5a)^{2}}- 5)(x-a)^{1}\\&x=10\\&a=\frac{\pi }{2}\\&f(10)\approx-42.15\end{align*}$

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Example Question #13 : Taylor Polynomial Approximation

$\begin{align*}&\text{For the function: }f(x)=cos(x) +\frac{ 1}{cos(8x)}\\&\text{Approximate }f(5)\\&\text{Using a Taylor expansion centered at }-4\cdot \pi\text{ to }3\text{ terms}\end{align*}$

Possible Answers:

19442.37

1080.24

-19442.37

9722.19

Correct answer:

9722.19

Explanation:

$\begin{align*}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function }cos(x) +\frac{ 1}{cos(8x)}\text{ at }x=5\\&\text{This expansion becomes:}\\&f(x)\approx (cos(a) +\frac{ 1}{cos(8a)})+(\frac{(8sin(8a))}{cos(8a)^{2}}- sin(a))(x-a)^{1}+\frac{\frac{64}{cos(8a)}- cos(a) +\frac{ (128sin(8a)^{2})}{cos(8a)^{3}}}{2}(x-a)^{2}\\&x=5\\&a=-4\cdot \pi\\&f(5)\approx9722.19\end{align*}$

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Example Question #18 : Ap Calculus Bc

$\begin{align*}&\text{For the function: }f(x)=-cos(6x)sin(2x)\\&\text{Approximate }f(9)\\&\text{Using a Taylor expansion centered at }-\frac{(3\cdot \pi )}{2}\text{ to }2\text{ terms}\end{align*}$

Possible Answers:

-27.42

-4.57

27.42

-13.71

Correct answer:

-27.42

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function }-cos(6x)sin(2x)\text{ at }x=9\\&\text{This expansion becomes:}\\&f(x)\approx (-cos(6a)sin(2a))+(6sin(2a)sin(6a) - 2cos(2a)cos(6a))(x-a)^{1}\\&x=9\\&a=-\frac{(3\cdot \pi )}{2}\\&f(9)\approx-27.42\end{align*}$

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Example Question #19 : Ap Calculus Bc

$\begin{align*}&\text{Approximate }f(9) \\&\text{Where }f(x)=-\frac{cos(8x)}{cos(4x)}\\&\text{By using a Taylor expansion to }2\text{ terms, centered at }\frac{(5\cdot \pi )}{4}\end{align*}$

Possible Answers:

5.00

1.00

8.00

-617.65

Correct answer:

1.00

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function }-\frac{cos(8x)}{cos(4x)}\text{ at }x=9\\&\text{This expansion becomes:}\\&f(x)\approx (-\frac{cos(8a)}{cos(4a)})+(\frac{(8sin(8a))}{cos(4a)}-\frac{ (4cos(8a)sin(4a))}{cos(4a)^{2}})(x-a)^{1}\\&x=9\\&a=\frac{(5\cdot \pi )}{4}\\&f(9)\approx1.00\end{align*}$

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Example Question #14 : Taylor Polynomial Approximation

$\begin{align*}&\text{For the function: }f(x)=sin(2x) - cos(7x)\\&\text{Approximate }f(7)\\&\text{Using a Taylor expansion centered at }-2\cdot \pi\text{ to }3\text{ terms}\end{align*}$

Possible Answers:

8671.27

1224.45

4295.29

4348.42

Correct answer:

4348.42

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function }sin(2x) - cos(7x)\text{ at }x=7\\&\text{This expansion becomes:}\\&f(x)\approx (sin(2a) - cos(7a))+(2cos(2a) + 7sin(7a))(x-a)^{1}+\frac{49cos(7a) - 4sin(2a)}{2}(x-a)^{2}\\&x=7\\&a=-2\cdot \pi\\&f(7)\approx4348.42\end{align*}$

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AP Calculus BC : AP Calculus BC

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #11 : Ap Calculus Bc

Example Question #12 : Ap Calculus Bc

Example Question #13 : Ap Calculus Bc

Example Question #14 : Ap Calculus Bc

Example Question #11 : Taylor Polynomial Approximation

Example Question #12 : Taylor Polynomial Approximation

Example Question #13 : Taylor Polynomial Approximation

Example Question #18 : Ap Calculus Bc

Example Question #19 : Ap Calculus Bc

Example Question #14 : Taylor Polynomial Approximation

All AP Calculus BC Resources

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