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AP Calculus BC : AP Calculus BC

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Example Questions

Example Question #51 : Polynomial Approximations And 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

196

388

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 #51 : Taylor 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:

$-\frac{352}{3}$

$\frac{212}{3}$

$\frac{152}{3}$

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:

256

$\frac{469}{6}$

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 #51 : Ap Calculus Bc

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

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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Example Question #51 : Polynomial Approximations And Series

$\begin{align*}&\text{Approximate }f(-2) \\&\text{Where }f(0)=8;f'(0)=-4;f''(0)=20\\&\text{By using a Maclaurin expansion.}\end{align*}$

Possible Answers:

$-\frac{152}{3}$

Correct answer:

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function values }f(0)=8;f'(0)=-4;f''(0)=20\\&\text{ at }x=-2\\&\text{This expansion becomes:}\\&f(-2)\approx \frac{8}{0!}(-2)^{0}+\frac{-4}{1!}(-2)^{1}+\frac{20}{2!}(-2)^{2}\\&f(-2)\approx56\end{align*}$

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Example Question #52 : Polynomial Approximations And Series

$\begin{align*}&\text{Approximate }f(3) \\&\text{Where }f(0)=-5;f'(0)=-14;f''(0)=11\\&\text{With a Maclaurin series expansion.}\end{align*}$

Possible Answers:

$\frac{5}{2}$

$-\frac{81}{2}$

$-\frac{57}{2}$

Correct answer:

$\frac{5}{2}$

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function values }f(0)=-5;f'(0)=-14;f''(0)=11\\&\text{ at }x=3\\&\text{This expansion becomes:}\\&f(3)\approx \frac{-5}{0!}(3)^{0}+\frac{-14}{1!}(3)^{1}+\frac{11}{2!}(3)^{2}\\&f(3)\approx\frac{5}{2}\end{align*}$

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Example Question #53 : Polynomial Approximations And Series

$\begin{align*}&\text{For the function values: }\\&f(0)=-8;f'(0)=1;f''(0)=14;f'''(0)=4;f''''(0)=-7\\&\text{Approximate }f(-3)\text{ using a Maclaurin expansion.}\end{align*}$

Possible Answers:

$-\frac{273}{40}$

$\frac{83}{8}$

$-\frac{9}{8}$

Correct answer:

$\frac{83}{8}$

Explanation:

$\begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Similarly, it is used to approximate an unknown function value,}\\&\text{if values of the function and its derivatives are known at a}\\&\text{particular point. In the case of the Maclaurin series, this}\\&\text{ point has the value of zero (for many functions f(0) is readily}\\&\text{known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function values }f(0)=-8;f'(0)=1;f''(0)=14;f'''(0)=4;f''''(0)=-7\\&\text{ at }x=-3\\&\text{This expansion becomes:}\\&f(-3)\approx \frac{-8}{0!}(-3)^{0}+\frac{1}{1!}(-3)^{1}+\frac{14}{2!}(-3)^{2}+\frac{4}{3!}(-3)^{3}+\frac{-7}{4!}(-3)^{4}\\&f(-3)\approx\frac{83}{8}\end{align*}$

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Example Question #51 : Ap Calculus Bc

Write out the first three terms of the Taylor series about a=1 for the following function:

f(x)=x^2+e^x

Possible Answers:

0+(1+e^1)+[(2+e^1)(x-1)]

$(1+e^1)+[(2+e^1)(x-1)]+\frac{(2+e^1)(x-1)^2}{2}$

(1+e^1)+[(2+e^1)(x-1)]+(x-1)^2

0+0+(x-1)^2

Correct answer:

$(1+e^1)+[(2+e^1)(x-1)]+\frac{(2+e^1)(x-1)^2}{2}$

Explanation:

The Taylor series about x=a for a function is given by

$\sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}$

For the first three terms (n=0, 1, 2) we must find the zeroth, first, and second derivative of the function. The zeroth derivative is just the function itself.

f'(x)=2x+e^x

f''(x)=2+e^x

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$

Now, using the above formula, write out the first three terms:

$\frac{(1+e^1)(x-1)^0}{0!}+\frac{[(2+e^1)(x-1)^1]}{1!}+\frac{(2+e^1)(x-1)^2}{2!}$

which simplified becomes

$(1+e^1)+[(2+e^1)(x-1)]+\frac{(2+e^1)(x-1)^2}{2}$

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Example Question #52 : Taylor Series

Find the Maclaurin Series of the function

$f(x)=sin\,x$

up to the fifth degree.

Possible Answers:

DNE

$1-\frac{x^2}{2!}+\frac{x^4}{4!}$

$x+\frac{x^2}{2!}-\frac{x^3}{3!}-\frac{x^4}{4!}+\frac{x^5}{5!}$

$x-\frac{x^3}{3!}+\frac{x^5}{5!}$

Correct answer:

$x-\frac{x^3}{3!}+\frac{x^5}{5!}$

Explanation:

The formula for an i-th degree Maclaurin Polynomial is

$f(x)=\sum_{n=0}^{i}\frac{f^{(n)}(0)}{n!}x^n$

For the fifth degree polynomial, we must evaluate the function up to its fifth derivative.

$\begin{matrix} f(0)=sin(0)=0\\ f'(0)=cos(0)=1\\ f''(0)=-sin(0)=0\\ f'''(0)=-cos(0)=-1\\ f^{(4)}=sin(0)=0 \\ \end{matrix}$

$f^{(5)}(0)=cos(0)=1$

The summation becomes

$\sum_{n=0}^{5}\frac{f^{(n)}(0)}{n!}x^n=\frac{f(0)}{0!}x^0+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\frac{f^{(4)}(0)}{4!}x^4+\frac{f^{(5)}(0)}{5!}x^5$

And substituting for the values of the function and the first five derivative values, we get the Maclaurin Polynomial

$f(x)=sin(x)\approx x-\frac{x^3}{3!}+\frac{x^5}{5!}$

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Example Question #53 : Taylor Series

Which of following intervals of convergence cannot exist?

Possible Answers:

For any $\small \epsilon>0$ , the interval $\small [a-\epsilon,a+\epsilon]$ for some $\small a\in\mathbb{R}$

$\small [2,10000]$

$\small [2,\infty)$

For any $\small q,p$ such that $\small q \leq p$ , the interval [q,p]

Correct answer:

$\small [2,\infty)$

Explanation:

$\small [2,\infty)$ cannot be an interval of convergence because a theorem states that a radius has to be either nonzero and finite, or infinite (which would imply that it has interval of convergence $\small (-\infty,\infty)$ ). Thus, $\small [2,\infty)$ can never be an interval of convergence.

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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 #51 : Polynomial Approximations And Series

Example Question #51 : Taylor Series

Example Question #29 : Maclaurin Series

Example Question #51 : Ap Calculus Bc

Example Question #51 : Polynomial Approximations And Series

Example Question #52 : Polynomial Approximations And Series

Example Question #53 : Polynomial Approximations And Series

Example Question #51 : Ap Calculus Bc

Example Question #52 : Taylor Series

Example Question #53 : Taylor Series

All AP Calculus BC Resources

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