Finding Taylor polynomial approximations of functions — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: This chapter covers definition of Taylor polynomials, nth-degree Taylor polynomials centered at a point, Maclaurin polynomials, coefficient derivation, and manipulation-based construction of polynomial approximations for sufficiently differentiable functions.

You should already know: How to compute higher-order derivatives, how to evaluate derivatives at a point, basic polynomial algebra.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Finding Taylor polynomial approximations of functions?

A Taylor polynomial is a polynomial approximation of a general non-polynomial function built to match the original function’s value and first $n$ derivatives at a specific point called the center. Unlike first-degree tangent line (linear) approximation, higher-degree Taylor polynomials match more derivatives, producing a much more accurate approximation over a larger interval around the center. When centered at $x=0$, Taylor polynomials are called Maclaurin polynomials, a common special case tested heavily on the exam.

According to the AP Calculus BC CED, this topic is part of Unit 10 (Infinite Sequences and Series), which accounts for 17-18% of total exam score, with this subtopic contributing roughly 2-4% of total points. This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections: MCQ typically asks for a specific coefficient or full nth-degree polynomial, while FRQ often combines polynomial construction with approximation or error analysis.

2. nth-Degree Taylor Polynomial: Definition and Coefficient Derivation

If a function $f$ is $n$ times differentiable at a point $x=a$, the $n$th-degree Taylor polynomial of $f$ centered at $a$ is the unique polynomial that matches $f$ and its first $n$ derivatives at $x=a$. To derive the formula, we start with the general form of a polynomial centered at $a$: $$P_n(x)=c_0+c_1(x-a)+c_2(x-a{)}^2+...+c_n(x-a{)}^n$$ To match $f(a)=P_n(a)$, plug $x=a$: all terms except $c_0$ vanish, so $c_0=f(a)$. Take the first derivative: $P_n^{\prime }(x)=c_1+2c_2(x-a)+...+n{c}_n(x-a{)}^{n-1}$, so $P_n^{\prime }(a)=c_1=f^{\prime }(a)$. Repeating this for the $k$th derivative gives $P_n^{(k)}(a)=k!\cdot c_k=f^{(k)}(a)$, so rearranging gives the general coefficient formula $c_k=\frac{f^{(k)}(a)}{k!}$, where $f^{(k)}(a)$ is the $k$th derivative of $f$ at $a$, $0!=1$, and $f^{(0)}(a)=f(a)$ by convention. The full Taylor polynomial formula is: $$P_n(x)=\sum _{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a{)}^k$$ This formula ensures the polynomial behaves identically to $f$ near $x=a$; higher $n$ gives more matched derivatives and a more accurate approximation.

Worked Example

Problem: Find the 3rd-degree Taylor polynomial of $f(x)=x^4-2x^3+5$ centered at $a=1$.

Solution:

1. Compute $f$ and its first 3 derivatives, evaluate each at $a=1$:
   • $f(x)=x^4-2x^3+5⟹f(1)=1-2+5=4$
   • $f^{\prime }(x)=4x^3-6x^2⟹f^{\prime }(1)=4-6=-2$
   • $f^{\prime \prime }(x)=12x^2-12x⟹f^{\prime \prime }(1)=12-12=0$
   • $f^{\prime \prime \prime }(x)=24x-12⟹f^{\prime \prime \prime }(1)=24-12=12$
2. Calculate each coefficient using $c_k=f^{(k)}(a)/k!$:
   • $c_0=4/0!=4$, $c_1=-2/1!=-2$, $c_2=0/2!=0$, $c_3=12/3!=2$
3. Substitute into the Taylor polynomial formula: $$P_3(x)=4-2(x-1)+0(x-1{)}^2+2(x-1{)}^3$$
4. If required, simplify to standard polynomial form: $P_3(x)=2x^3-8x^2+10x$, which matches the original function truncated to degree 3, as expected.

Exam tip: Always write each derivative and its evaluation at the center explicitly step by step. AP exam readers award partial credit for correct intermediate derivatives even if your final coefficient is wrong, so never skip these steps.

3. Maclaurin Polynomials (Taylor Polynomials Centered at 0)

A Maclaurin polynomial is simply a Taylor polynomial with center $a=0$, the most common type of Taylor polynomial on the AP exam. The formula simplifies directly from the general Taylor formula by substituting $a=0$: $$P_n(x)=\sum _{k=0}^n\frac{f^{(k)}(0)}{k!}{x}^k$$ Because $(x-a)=x-0=x$, the $(x-a{)}^k$ term reduces to $x^k$, making calculations faster. Maclaurin polynomials are used to approximate functions near $x=0$, which is often the point of interest for applications, and standard Maclaurin polynomials for common functions (like $e^x$, $\sin x$, $\cos x$) are frequently tested. The process for constructing a Maclaurin polynomial is identical to the general Taylor process, only the evaluation point for derivatives changes.

Worked Example

Problem: Find the 4th-degree Maclaurin polynomial for $f(x)=\sin (2x)$.

Solution:

1. Compute $f$ and its first 4 derivatives, evaluate each at $a=0$:
   • $f(x)=\sin (2x)⟹f(0)=\sin (0)=0$
   • $f^{\prime }(x)=2\cos (2x)⟹f^{\prime }(0)=2\cos (0)=2$
   • $f^{\prime \prime }(x)=-4\sin (2x)⟹f^{\prime \prime }(0)=-4\sin (0)=0$
   • $f^{\prime \prime \prime }(x)=-8\cos (2x)⟹f^{\prime \prime \prime }(0)=-8\cos (0)=-8$
   • $f^{(4)}(x)=16\sin (2x)⟹f^{(4)}(0)=16\sin (0)=0$
2. Calculate coefficients:
   • $c_0=0/0!=0$, $c_1=2/1!=2$, $c_2=0/2!=0$, $c_3=-8/6=-4/3$, $c_4=0/24=0$
3. Substitute into the Maclaurin formula: $$P_4(x)=0+2x+0x^2-\frac{4}{3}{x}^3+0x^4=2x-\frac{4}{3}{x}^3$$ Because all even derivatives of $\sin (2x)$ at 0 are zero, the 4th-degree polynomial matches the 3rd-degree polynomial, which is expected for this odd function.

Exam tip: If you are asked for an nth-degree Maclaurin polynomial, you only need to include all terms up to degree $n$; if higher coefficients are zero, explicitly note this (or write the zero term) to show you completed the requirement.

4. Constructing Taylor Polynomials by Manipulation

Once you know the Taylor polynomial of $f(x)$ centered at $a$, you can find the Taylor polynomial of related transformed functions (like $f(kx)$, $f(x^m)$, $xf(x)$) by substituting, scaling, or differentiating the known polynomial, instead of computing all derivatives from scratch. This technique saves significant time on the AP exam, especially for multiple-choice questions where speed is critical. Common valid manipulations include:

1. Substitution: If $P_n(u)$ is the nth-degree Maclaurin polynomial for $f(u)$, then $P_n(g(x))$ is the Maclaurin polynomial for $f(g(x))$, up to degree $n$.
2. Scaling: The Taylor polynomial for $cf(x)$ is $c$ times the Taylor polynomial of $f(x)$, term by term.
3. Differentiation/integration: The derivative of the Taylor polynomial of $f(x)$ is the Taylor polynomial of $f^{\prime }(x)$, with the same center and degree $n-1$. When using manipulation, always discard any terms that become higher degree than the requested $n$ after substitution or multiplication.

Worked Example

Problem: The 5th-degree Maclaurin polynomial for $e^u$ is given as $P_5(u)=1+u+\frac{u^2}{2!}+\frac{u^3}{3!}+\frac{u^4}{4!}+\frac{u^5}{5!}$. Use this to find the 4th-degree Maclaurin polynomial for $f(x)=x{e}^{-2x^2}$.

Solution:

1. Substitute $u=-2x^2$ into the given polynomial for $e^u$: $$e^{-2x^2}\approx 1+(-2x^2)+\frac{(-2x^2{)}^2}{2!}+\frac{(-2x^2{)}^3}{3!}+...$$
2. We only need terms up to degree 4 after multiplying by $x$, so discard any terms that will become higher than degree 4: the $(-2x^2{)}^3$ term is degree 6, which becomes degree 7 after multiplying by $x$, so we discard it. Simplify the remaining terms: $$e^{-2x^2}\approx 1-2x^2+\frac{4x^4}{2}=1-2x^2+2x^4$$
3. Multiply by $x$ and keep only terms up to degree 4: $$x{e}^{-2x^2}\approx x(1-2x^2+2x^4)=x-2x^3+2x^5$$
4. The $2x^5$ term is degree 5, which is higher than the requested 4, so we drop it. The final polynomial is: $$P_4(x)=x-2x^3$$

Exam tip: When substituting, always check the degree of each term after substitution before writing your final answer. It is extremely common for higher-degree terms from the original polynomial to become higher than the requested degree after substitution, so always double-check.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting to divide by $k!$ when calculating the coefficient of $(x-a{)}^k$. Why: Students confuse the Taylor polynomial formula with general power series, or forget the derivation step that introduces the factorial. Correct move: Every time you write a coefficient, explicitly write $\frac{f^{(k)}(a)}{k!}$ before simplifying, even if $k!=1$ for $k=0$ or $k=1$.
• Wrong move: Evaluating derivatives at 0 for a Taylor polynomial with center $a\ne 0$. Why: Students get used to Maclaurin polynomials and automatically evaluate at 0 by habit. Correct move: Immediately after reading the problem, write "Center $a=[givenvalue]$" at the top of your work, and check the evaluation point every time you compute a derivative.
• Wrong move: Forgetting to apply the chain rule when computing derivatives of composite functions. Why: Students rush derivative calculations and miss the inner function derivative, leading to incorrect coefficients. Correct move: After computing each derivative, pause and confirm you applied the chain rule before evaluating at the center.
• Wrong move: Keeping higher-degree terms after substitution that exceed the requested degree $n$. Why: Students substitute all terms from the original polynomial without checking the final degree after transformation. Correct move: After manipulation, sort terms by degree and drop any term with degree strictly greater than the requested $n$.
• Wrong move: Incorrectly expanding $(x-a{)}^k=x^k-a^k$. Why: Students make a basic algebra mistake when converting to standard polynomial form. Correct move: Expand $(x-a{)}^k$ step-by-step or use the binomial theorem, never split the exponent across subtraction.
• Wrong move: Stopping at the last non-zero term and not including all terms up to degree $n$. Why: Students think that nth-degree means n non-zero terms, but the AP exam requires all terms up to $n$. Correct move: Always write out all terms from degree 0 up to degree $n$, writing a 0 coefficient if needed, to show you completed the requirement.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Which of the following is the coefficient of $(x-2{)}^3$ in the 3rd-degree Taylor polynomial of $f(x)=\ln x$ centered at $a=2$? A) $\frac{1}{24}$ B) $-\frac{1}{12}$ C) $\frac{1}{8}$ D) $-\frac{1}{24}$

Worked Solution: To find the coefficient of $(x-a{)}^k$ in a Taylor polynomial, we use the formula $c_k=\frac{f^{(k)}(a)}{k!}$. First compute the third derivative of $f(x)=\ln x$: $f^{\prime }(x)=x^{-1}$, $f^{\prime \prime }(x)=-x^{-2}$, $f^{\prime \prime \prime }(x)=2x^{-3}$. Evaluate the third derivative at $a=2$: $f^{\prime \prime \prime }(2)=2(2^{-3})=\frac{2}{8}=\frac{1}{4}$. Divide by $3!=6$ to get the coefficient: $c_3=\frac{1/4}{6}=\frac{1}{24}$. The correct answer is A.

Question 2 (Free Response)

Let $f(x)=\frac{1}{1+x}$. (a) Find the 4th-degree Maclaurin polynomial for $f(x)$. (b) Use the polynomial from part (a) to approximate $\frac{1}{1.2}$. Round your approximation to 4 decimal places. (c) Find the 3rd-degree Taylor polynomial for $f(x)$ centered at $a=1$.

Worked Solution: (a) First compute derivatives and evaluate at 0: $f(x)=(1+x{)}^{-1}⟹f(0)=1$, $f^{\prime }(x)=-(1+x{)}^{-2}⟹f^{\prime }(0)=-1$, $f^{\prime \prime }(x)=2(1+x{)}^{-3}⟹f^{\prime \prime }(0)=2$, $f^{\prime \prime \prime }(x)=-6(1+x{)}^{-4}⟹f^{\prime \prime \prime }(0)=-6$, $f^{(4)}(x)=24(1+x{)}^{-5}⟹f^{(4)}(0)=24$. Coefficients: $c_k=f^{(k)}(0)/k!$, so $c_0=1,c_1=-1,c_2=1,c_3=-1,c_4=1$. The 4th-degree Maclaurin polynomial is $P_4(x)=1-x+x^2-x^3+x^4$.

(b) $\frac{1}{1.2}=\frac{1}{1+0.2}$, so substitute $x=0.2$: $P_4(0.2)=1-0.2+(0.2{)}^2-(0.2{)}^3+(0.2{)}^4=0.8336$. The approximation is 0.8336.

(c) Evaluate derivatives at $a=1$: $f(1)=1/2$, $f^{\prime }(1)=-1/4$, $f^{\prime \prime }(1)=1/4$, $f^{\prime \prime \prime }(1)=-3/8$. Coefficients: $c_0=1/2$, $c_1=-1/4$, $c_2=1/8$, $c_3=-1/16$. The 3rd-degree Taylor polynomial is $P_3(x)=\frac{1}{2}-\frac{1}{4}(x-1)+\frac{1}{8}(x-1{)}^2-\frac{1}{16}(x-1{)}^3$.

Question 3 (Application / Real-World Style)

The position of a particle moving along the x-axis at time $t$ (seconds, $0\le t\le 1$) is given by $s(t)=e^{\sin t}$, where $s(t)$ is measured in meters. For small values of $t$ near 0, physicists use a 3rd-degree Taylor polynomial approximation of $s(t)$ to simplify calculations. Find the 3rd-degree Maclaurin polynomial for $s(t)$, then use it to approximate the particle's position at $t=0.5$ seconds.

Worked Solution: First compute $s(t)$ and its first 3 derivatives evaluated at $t=0$: $s(t)=e^{\sin t}⟹s(0)=1$, $s^{\prime }(t)=\cos t\cdot e^{\sin t}⟹s^{\prime }(0)=1$, $s^{\prime \prime }(t)=-\sin t{e}^{\sin t}+{\cos }^{2}t{e}^{\sin t}⟹s^{\prime \prime }(0)=1$, $s^{\prime \prime \prime }(t)=-\cos t{e}^{\sin t}-\sin t\cos t{e}^{\sin t}-2\sin t\cos t{e}^{\sin t}+{\cos }^{3}t{e}^{\sin t}⟹s^{\prime \prime \prime }(0)=0$. Coefficients: $c_0=1$, $c_1=1$, $c_2=1/2!=1/2$, $c_3=0/3!=0$. The 3rd-degree Maclaurin polynomial is $P_3(t)=1+t+\frac{1}{2}{t}^2$. Evaluate at $t=0.5$: $P_3(0.5)=1+0.5+\frac{1}{2}(0.5{)}^2=1.625$ meters. In context, this approximation means the particle is approximately 1.625 meters from the origin at 0.5 seconds.

7. Quick Reference Cheatsheet

8. What's Next

Mastering Taylor polynomial construction is the non-negotiable foundation for all remaining topics in Unit 10, Infinite Sequences and Series. Without being able to correctly find Taylor coefficients and build nth-degree approximations, you cannot work with infinite Taylor series, test convergence, or calculate approximation error, all of which are heavily weighted on the AP Calculus BC exam. This topic specifically prepares you for Taylor's theorem and Lagrange error bound, a common FRQ topic that requires you to use your constructed polynomial to bound the error of your approximation. Next, you will extend the finite nth-degree Taylor polynomial to an infinite Taylor series, which lets you represent non-polynomial functions as infinite power series, enabling calculations like integrating non-elementary functions that cannot be done with basic integration rules.

• Taylor series
• Radius and interval of convergence for power series
• Lagrange error bound for Taylor approximations