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AP Calculus BC : AP Calculus BC

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2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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Example Question #1 : Polynomial Approximations And Series

$\begin{align*}&\text{Approximate }f(-4) \\&\text{Where }f(-3)=-3;f'(-3)=-13;f''(-3)=-10;f'''(-3)=-20\\&\text{With a Taylor series expansion.}\end{align*}$

Possible Answers:

$\frac{25}{3}$

$-\frac{73}{3}$

$-\frac{8}{3}$

Correct answer:

$\frac{25}{3}$

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function values }f(-3)=-3;f'(-3)=-13;f''(-3)=-10;f'''(-3)=-20\\&\text{ at }x=-4\\&\text{This expansion becomes:}\\&f(-4)\approx \frac{-3}{0!}(-4-(-3))^{0}+\frac{-13}{1!}(-4-(-3))^{1}+\frac{-10}{2!}(-4-(-3))^{2}+\frac{-20}{3!}(-4-(-3))^{3}\\&f(-4)\approx\frac{25}{3}\end{align*}$

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Example Question #1 : Ap Calculus Bc

$\begin{align*}&\text{Approximate }f(-7) \\&\text{Where }f(-2)=7;f'(-2)=-19;f''(-2)=13;f'''(-2)=10;f''''(-2)=-16\\&\text{By using a Taylor expansion.}\end{align*}$

Possible Answers:

$\frac{535}{4}$

$-\frac{721}{2}$

$-\frac{803}{6}$

Correct answer:

$-\frac{721}{2}$

Explanation:

$\begin{align*}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function values }f(-2)=7;f'(-2)=-19;f''(-2)=13;f'''(-2)=10;f''''(-2)=-16\\&\text{ at }x=-7\\&\text{This expansion becomes:}\\&f(-7)\approx \frac{7}{0!}(-7-(-2))^{0}+\frac{-19}{1!}(-7-(-2))^{1}+\frac{13}{2!}(-7-(-2))^{2}+\frac{10}{3!}(-7-(-2))^{3}+\frac{-16}{4!}(-7-(-2))^{4}\\&f(-7)\approx-\frac{721}{2}\end{align*}$

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Example Question #1 : Taylor Series

$\begin{align*}&\text{Approximate }f(7) \\&\text{Where }f(3)=7;f'(3)=16;f''(3)=1\\&\text{With a Taylor series expansion.}\end{align*}$

Possible Answers:

$\frac{500}{3}$

Correct answer:

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function values }f(3)=7;f'(3)=16;f''(3)=1\\&\text{ at }x=7\\&\text{This expansion becomes:}\\&f(7)\approx \frac{7}{0!}(7-3)^{0}+\frac{16}{1!}(7-3)^{1}+\frac{1}{2!}(7-3)^{2}\\&f(7)\approx79\end{align*}$

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Example Question #1 : Ap Calculus Bc

$\begin{align*}&\text{For the function values: }\\&f(9)=10;f'(9)=-16;f''(9)=7;f'''(9)=-2;f''''(9)=-12\\&\text{Approximate }f(6)\text{ using a Taylor expansion.}\end{align*}$

Possible Answers:

$-\frac{2319}{20}$

Correct answer:

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function values }f(9)=10;f'(9)=-16;f''(9)=7;f'''(9)=-2;f''''(9)=-12\\&\text{ at }x=6\\&\text{This expansion becomes:}\\&f(6)\approx \frac{10}{0!}(6-9)^{0}+\frac{-16}{1!}(6-9)^{1}+\frac{7}{2!}(6-9)^{2}+\frac{-2}{3!}(6-9)^{3}+\frac{-12}{4!}(6-9)^{4}\\&f(6)\approx58\end{align*}$

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Example Question #1 : Taylor Series

$\begin{align*}&\text{For the function values: }\\&f(-12)=6;f'(-12)=19;f''(-12)=18;f'''(-12)=-2\\&\text{Approximate }f(-10)\text{ using a Taylor expansion.}\end{align*}$

Possible Answers:

$\frac{232}{3}$

$\frac{20}{3}$

$\frac{218}{3}$

Correct answer:

$\frac{232}{3}$

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function values }f(-12)=6;f'(-12)=19;f''(-12)=18;f'''(-12)=-2\\&\text{ at }x=-10\\&\text{This expansion becomes:}\\&f(-10)\approx \frac{6}{0!}(-10-(-12))^{0}+\frac{19}{1!}(-10-(-12))^{1}+\frac{18}{2!}(-10-(-12))^{2}+\frac{-2}{3!}(-10-(-12))^{3}\\&f(-10)\approx\frac{232}{3}\end{align*}$

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Example Question #1 : Ap Calculus Bc

$\begin{align*}&\text{For the function values: }\\&f(-2)=2;f'(-2)=10;f''(-2)=-3;f'''(-2)=-3\\&\text{Approximate }f(-1)\text{ using a Taylor expansion.}\end{align*}$

Possible Answers:

$\frac{51}{8}$

Correct answer:

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function values }f(-2)=2;f'(-2)=10;f''(-2)=-3;f'''(-2)=-3\\&\text{ at }x=-1\\&\text{This expansion becomes:}\\&f(-1)\approx \frac{2}{0!}(-1-(-2))^{0}+\frac{10}{1!}(-1-(-2))^{1}+\frac{-3}{2!}(-1-(-2))^{2}+\frac{-3}{3!}(-1-(-2))^{3}\\&f(-1)\approx10\end{align*}$

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Example Question #1 : Taylor Series

$\begin{align*}&\text{Approximate }f(3) \\&\text{Where }f(8)=17;f'(8)=12;f''(8)=3\\&\text{By using a Taylor expansion.}\end{align*}$

Possible Answers:

$-\frac{11}{2}$

$\frac{229}{2}$

152

$\frac{5}{2}$

Correct answer:

$-\frac{11}{2}$

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function values }f(8)=17;f'(8)=12;f''(8)=3\\&\text{ at }x=3\\&\text{This expansion becomes:}\\&f(3)\approx \frac{17}{0!}(3-8)^{0}+\frac{12}{1!}(3-8)^{1}+\frac{3}{2!}(3-8)^{2}\\&f(3)\approx-\frac{11}{2}\end{align*}$

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Example Question #1 : Polynomial Approximations And Series

$\begin{align*}&\text{Approximate }f(-5) \\&\text{Where }f(-2)=-18;f'(-2)=11;f''(-2)=7\\&\text{With a Taylor series expansion.}\end{align*}$

Possible Answers:

$\frac{93}{2}$

$-\frac{39}{2}$

Correct answer:

$-\frac{39}{2}$

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function values }f(-2)=-18;f'(-2)=11;f''(-2)=7\\&\text{ at }x=-5\\&\text{This expansion becomes:}\\&f(-5)\approx \frac{-18}{0!}(-5-(-2))^{0}+\frac{11}{1!}(-5-(-2))^{1}+\frac{7}{2!}(-5-(-2))^{2}\\&f(-5)\approx-\frac{39}{2}\end{align*}$

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Example Question #1 : Taylor Series

$\begin{align*}&\text{Approximate }f(3) \\&\text{Where }f(2)=13;f'(2)=-7;f''(2)=13;f'''(2)=12\\&\text{With a Taylor series expansion.}\end{align*}$

Possible Answers:

$\frac{73}{6}$

$\frac{29}{2}$

$\frac{49}{2}$

Correct answer:

$\frac{29}{2}$

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function values }f(2)=13;f'(2)=-7;f''(2)=13;f'''(2)=12\\&\text{ at }x=3\\&\text{This expansion becomes:}\\&f(3)\approx \frac{13}{0!}(3-2)^{0}+\frac{-7}{1!}(3-2)^{1}+\frac{13}{2!}(3-2)^{2}+\frac{12}{3!}(3-2)^{3}\\&f(3)\approx\frac{29}{2}\end{align*}$

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Example Question #1 : Taylor Series

$\begin{align*}&\text{Approximate }f(6) \\&\text{Where }f(11)=-18;f'(11)=15;f''(11)=5;f'''(11)=-6\\&\text{With a Taylor series expansion.}\end{align*}$

Possible Answers:

$-\frac{11}{2}$

$\frac{189}{2}$

$\frac{205}{12}$

Correct answer:

$\frac{189}{2}$

Explanation:

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\\&\text{at given point, if values for the function and its derivatives}\\&\text{are known at another point. It takes advantage of rates of change}\\&\text{(i.e. function derivative values), and the distance between}\\&\text{points on the x-axis to approximate changes along the y-axis.}\\&\text{It is similar to using the slope of a line to find changes in}\\&\text{y with regards to changes in x. The Taylor series follows the}\\&\text{form:}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\\&\text{For the function values }f(11)=-18;f'(11)=15;f''(11)=5;f'''(11)=-6\\&\text{ at }x=6\\&\text{This expansion becomes:}\\&f(6)\approx \frac{-18}{0!}(6-11)^{0}+\frac{15}{1!}(6-11)^{1}+\frac{5}{2!}(6-11)^{2}+\frac{-6}{3!}(6-11)^{3}\\&f(6)\approx\frac{189}{2}\end{align*}$

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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 #1 : Polynomial Approximations And Series

Example Question #1 : Ap Calculus Bc

Example Question #1 : Taylor Series

Example Question #1 : Ap Calculus Bc

Example Question #1 : Taylor Series

Example Question #1 : Ap Calculus Bc

Example Question #1 : Taylor Series

Example Question #1 : Polynomial Approximations And Series

Example Question #1 : Taylor Series

Example Question #1 : Taylor Series

All AP Calculus BC Resources

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