We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

AP Calculus BC : Taylor Polynomial Approximation

Study concepts, example questions & explanations for AP Calculus BC

CREATE AN ACCOUNT Create Tests & Flashcards

Home Embed

All AP Calculus BC Resources

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

Example Questions

← Previous 1 2 3 Next →

Example Question #1 : Taylor Polynomial Approximation

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

Possible Answers:

$-\frac{73}{3}$

$-\frac{8}{3}$

$\frac{25}{3}$

Correct answer:

$\frac{25}{3}$

Explanation:

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

Report an Error

Example Question #1 : Taylor Polynomial Approximation

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

Possible Answers:

$-\frac{803}{6}$

$\frac{535}{4}$

$-\frac{721}{2}$

Correct answer:

$-\frac{721}{2}$

Explanation:

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

Report an Error

Example Question #1 : Ap Calculus Bc

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

Possible Answers:

$\frac{500}{3}$

Correct answer:

Explanation:

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

Report an Error

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:

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

Report an Error

Example Question #1 : Ap Calculus Bc

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

Possible Answers:

$\frac{232}{3}$

$\frac{20}{3}$

$\frac{218}{3}$

Correct answer:

$\frac{232}{3}$

Explanation:

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

Report an Error

Example Question #2 : Taylor Series

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

Possible Answers:

$\frac{51}{8}$

Correct answer:

Explanation:

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

Report an Error

Example Question #1 : Polynomial Approximations And 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:

$\frac{5}{2}$

$-\frac{11}{2}$

152

$\frac{229}{2}$

Correct answer:

$-\frac{11}{2}$

Explanation:

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

Report an Error

Example Question #1 : Taylor Polynomial Approximation

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

Possible Answers:

$-\frac{39}{2}$

$\frac{93}{2}$

Correct answer:

$-\frac{39}{2}$

Explanation:

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

Report an Error

Example Question #1 : Taylor Polynomial Approximation

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

Possible Answers:

$\frac{29}{2}$

$\frac{49}{2}$

$\frac{73}{6}$

Correct answer:

$\frac{29}{2}$

Explanation:

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

Report an Error

Example Question #10 : 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:

$\frac{205}{12}$

$\frac{189}{2}$

$-\frac{11}{2}$

Correct answer:

$\frac{189}{2}$

Explanation:

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

Report an Error

← Previous 1 2 3 Next →

View AP Calculus BC Tutors

Jaspal
Certified Tutor

University of the Sciences, Bachelor of Science, Biology, General. Temple University, Master of Science, Statistics.

View AP Calculus BC Tutors

Yosef
Certified Tutor

Cooper Union for the Advancement of Science and Art, Bachelor of Engineering, Mechanical Engineering. Rensselaer Polytechnic ...

View AP Calculus BC Tutors

Gwynne
Certified Tutor

University of Hawaii at Manoa, Bachelor in Arts, Chemistry. University of Michigan-Ann Arbor, Master of Science, Medicinal an...

All AP Calculus BC Resources

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

Popular Subjects

Calculus Tutoring in Philadelphia, French Tutoring in Houston, Biology Tutoring in Boston, Chemistry Tutoring in Boston, English Tutoring in Washington DC, SAT Tutoring in Philadelphia, Calculus Tutoring in Seattle, Reading Tutoring in San Francisco-Bay Area, MCAT Tutoring in San Diego, Math Tutoring in Los Angeles

Popular Courses & Classes

SAT Courses & Classes in Miami, SSAT Courses & Classes in Los Angeles, GMAT Courses & Classes in Seattle, LSAT Courses & Classes in San Diego, ISEE Courses & Classes in Miami, ACT Courses & Classes in Atlanta, ISEE Courses & Classes in Dallas Fort Worth, Spanish Courses & Classes in Atlanta, SSAT Courses & Classes in Phoenix, SAT Courses & Classes in Dallas Fort Worth

Popular Test Prep

MCAT Test Prep in Atlanta, LSAT Test Prep in Philadelphia, ISEE Test Prep in Dallas Fort Worth, ACT Test Prep in Atlanta, SSAT Test Prep in Miami, MCAT Test Prep in Dallas Fort Worth, ISEE Test Prep in Phoenix, SSAT Test Prep in Washington DC, LSAT Test Prep in New York City, SAT Test Prep in Boston

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

AP Calculus BC : Taylor Polynomial Approximation

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #1 : Taylor Polynomial Approximation

Example Question #1 : Taylor Polynomial Approximation

Example Question #1 : Ap Calculus Bc

Example Question #1 : Ap Calculus Bc

Example Question #1 : Ap Calculus Bc

Example Question #2 : Taylor Series

Example Question #1 : Polynomial Approximations And Series

Example Question #1 : Taylor Polynomial Approximation

Example Question #1 : Taylor Polynomial Approximation

Example Question #10 : Taylor Series

All AP Calculus BC Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

FIND THE BEST TUTORS

Email address:
Your name:

Email address:
Your name:
Feedback: