A Taylor polynomial approximates a function near a point $x=a$ by matching the function's value and derivatives at that point. The coefficient of the $n$ th-degree term is $\frac{f^{(n)}(a)}{n!}$ , and a Taylor polynomial centered at $x=0$ is called a Maclaurin polynomial. For AP Calculus BC, organize values in a derivative table before building the polynomial.

How Do Taylor Polynomial Approximations Work?

A Taylor polynomial approximates $f(x)$ near a center $x=a$ by using the values of $f$ and its derivatives at $a$ . The $n$ th-degree term has coefficient $\frac{f^{(n)}(a)}{n!}$ and power $(x-a)^n$ , so each added term matches one more derivative at the center.

For AP Calculus BC, the core routine is: identify the center, compute the needed derivatives, evaluate them at the center, build the polynomial through the requested degree, and plug in a nearby $x$ -value if the problem asks for an approximation.

more resources to help you study

practice multiple choice FRQ practice & scoring cheatsheets score calculator key terms

Why This Matters for the AP Calculus Exam

Taylor polynomials are a core BC skill in Unit 10. They let you turn a complicated function like $e^{5x}$ , $\sin x$ , or $\ln(1+x)$ into a polynomial you can evaluate by hand. On the exam, you may need to build a Taylor or Maclaurin polynomial to a given degree, use one to approximate a function value near the center, or recognize the pattern in the coefficients. This topic also sets up later work with the Lagrange error bound and radius/interval of convergence, so building fluency here pays off across the whole unit.

Key Takeaways

The $n$ th-degree term coefficient in a Taylor polynomial centered at $x=a$ is $\frac{f^{(n)}(a)}{n!}$ .
A Maclaurin polynomial is just a Taylor polynomial centered at $x=0$ .
The $n$ th-degree Taylor polynomial is the $n$ th partial sum of the full Taylor series.
Higher-degree polynomials usually approximate the original function more accurately over an interval.
Use a derivative table ( $n$ , $f^{(n)}(x)$ , $f^{(n)}(a)$ , $(x-a)^n$ , full term) to stay organized.
Taylor polynomials are most accurate near the center $x=a$ and tend to drift away from $f$ as $x$ moves farther from $a$ .

The Taylor Polynomial Formula

For a function $f(x)$ , its Taylor series centered at $x=a$ is:

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

Written out term by term, this is:

f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n

Here $f^{(n)}(a)$ is the $n$ th derivative of the function evaluated at $a$ , and $f^{(0)}(a)=f(a)$ . The $n$ th-degree Taylor polynomial is the $n$ th partial sum of the infinite series, meaning you stop after the $(x-a)^n$ term.

Taylor series centered at $x=0$ are common enough to get their own name: Maclaurin series.

Breaking Down the Formula

Taylor series look intimidating at first. The trick is to organize the pieces in a table. Each row builds one term of the polynomial.

\begin{array}{ |c|c|c|c|c|c| } \hline n & n! & f^n(x) & f^n(a)&(x-a)^n & \frac{f^n(a)}{n!}\cdot(x-a)^n \\ 0 & 1 & f(x) & f(a)&(x-a)^0 & \frac{f(a)}{1}\cdot(x-a)^0 \\ 1 & 1 & f'(x) & f'(a)&(x-a)^1 & \frac{f'(a)}{1}\cdot(x-a)^1 \\ 2 & 2 & f''(x) & f''(a)&(x-a)^2 & \frac{f''(a)}{2}\cdot(x-a)^2 \\ 3 & 6 & f'''(x) & f'''(a)&(x-a)^3 & \frac{f'''(a)}{6}\cdot(x-a)^3 \\ ... & ... & ... & ... & ... & ... \\ n & n! & f^n(x) & f^n(a)&(x-a)^n & \frac{f^n(a)}{n!}\cdot(x-a)^n \\ \hline \end{array}

Fill in the table column by column, then add up the entries in the last column to get your polynomial.

Worked Example

Find the third-degree Maclaurin polynomial for $e^{5x}$ .

Solution: Build the table. A Maclaurin polynomial is just a Taylor polynomial with $a=0$ , so every derivative is evaluated at $0$ .

\begin{array}{ |c|c|c|c|c|c| } \hline n & n! & f^n(x) & f^n(a)&(x-a)^n & \frac{f^n(a)}{n!}\cdot(x-a)^n \\ 0 & 1 & e^{5x} & 1&1 & 1 \\ 1 & 1 & 5e^{5x} & 5&x & 5x \\ 2 & 2 & 25e^{5x} & 25&x^2 & 25x^2/2 \\ 3 & 6 & 125e^{5x} & 125&x^3 & 125x^3/6 \\ \hline \end{array}

Add the terms in the last column. The third-degree Maclaurin polynomial for $e^{5x}$ is:

1+5x+\frac{25}{2}x^2+\frac{125}{6}x^3

How to Use This on the AP Calculus Exam

Problem Solving

Identify the center $a$ first. If the problem says "Maclaurin," use $a=0$ .
Take derivatives one degree at a time and evaluate each at $a$ before moving on. It is easy to lose track if you do all the derivatives first.
Build the term using $\frac{f^{(n)}(a)}{n!}(x-a)^n$ . Keep the factorial in the denominator separate from any constant that comes from the derivative.
To approximate a function value near $a$ , plug the given $x$ into your finished polynomial. The closer $x$ is to $a$ , the more accurate the result.

Free Response

Show each derivative and its value at the center. Clear work makes your reasoning easy to follow.
Leave coefficients as exact fractions rather than rounding, unless the problem asks for a decimal approximation.
If a problem asks for a specific degree, stop at that degree. Do not add extra terms.

Common Trap

Watch the chain rule when taking repeated derivatives of functions like $e^{5x}$ or $\sin(2x)$ . Each derivative pulls down another factor.
For functions like $\ln(x)$ centered at $x=1$ , keep the $(x-1)^n$ form. Do not switch to powers of $x$ .

Practice

Try these on your own, then check the worked solutions.

Problems

Find the fifth-degree Maclaurin polynomial for $f(x)=\cos(x)$ .
Find the third-degree Taylor polynomial for $f(x)=\ln(x)$ about $x=1$ .
Find the fourth-degree Taylor polynomial about $x=2$ for $f(x)=\sqrt{x}$ .

Solution for Question 1

Build your table and fill in the values:

\begin{array}{ |c|c|c|c|c|c| } \hline n & n! & f^n(x) & f^n(a)&(x-a)^n & \frac{f^n(a)}{n!}\cdot(x-a)^n \\ 0 & 1 & \text{cos}(x) & 1&1 & 1 \\ 1 & 1 & -\text{sin}(x) & 0&x & 0 \\ 2 & 2 & -\text{cos}(x) & -1&x^2 & -x^2/2 \\ 3 & 6 & \text{sin}(x) & 0&x^3 & 0 \\ 4 & 24 & \text{cos}(x) & 1&x^4 & x^4/24 \\ 5 & 120 & -\text{sin}(x) & 0&x^5 & 0 \\ \hline \end{array}

Putting it together, the fifth-degree Maclaurin polynomial for $f(x)=\cos(x)$ is

1-\frac{x^2}{2}+\frac{x^4}{24}

Notice that the odd-degree terms all drop out because the sine values at $0$ are zero. This is why $\cos x$ has only even powers.

Solution for Question 2

Here the $(x-a)^n$ column uses $(x-1)$ since the center is $x=1$ .

\begin{array}{ |c|c|c|c|c|c| } \hline n & n! & f^n(x) & f^n(a)&(x-a)^n & \frac{f^n(a)}{n!}\cdot(x-a)^n \\ 0 & 1 & \text{ln}(x) & 0&1 & 0 \\ 1 & 1 & 1/x & 1&(x-1) & (x-1) \\ 2 & 2 & -1/x^2 & -1&(x-1)^2 & -\frac{1}{2}(x-1)^2 \\ 3 & 6 & 2/x^3 & 2&(x-1)^3 & \frac{1}{3}(x-1)^3 \\ \hline \end{array}

The third-degree term is $\frac{f'''(1)}{3!}(x-1)^3 = \frac{2}{6}(x-1)^3 = \frac{1}{3}(x-1)^3$ . The polynomial is:

(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3

Solution for Question 3

One more table, this time centered at $x=2$ .

\begin{array}{ |c|c|c|c|c|c| } \hline n & n! & f^n(x) & f^n(a)&(x-a)^n & \frac{f^n(a)}{n!}\cdot(x-a)^n \\ 0 & 1 & \sqrt{x} & \sqrt{2}&1 & \sqrt{2} \\ 1 & 1 & \frac{1}{2\sqrt{x}} & \frac{1}{2\sqrt{2}}&(x-2) & \frac{1}{2\sqrt{2}}\cdot(x-2)\\ 2 & 2 & -\frac{1}{4\sqrt{x^3}} & -\frac{1}{4\sqrt{8}}&(x-2)^2 & -\frac{1}{8\sqrt{8}}\cdot(x-2)^2 \\ 3 & 6 & \frac{3}{8\sqrt{x^5}} & \frac{3}{8\sqrt{32}}&(x-2)^3 & \frac{3}{48\sqrt{32}}\cdot (x-2)^3 \\ 4 & 24 & -\frac{15}{16\sqrt{x^7}} & -\frac{15}{16\sqrt{128}}&(x-2)^4 & -\frac{15}{384\sqrt{128}}\cdot(x-2)^4 \\ \hline \end{array}

Putting it all together:

\sqrt{2}+\frac{1}{2\sqrt{2}}\cdot(x-2)-\frac{1}{8\sqrt{8}}\cdot(x-2)^2+\frac{3}{48\sqrt{32}}\cdot (x-2)^3 -\frac{15}{384\sqrt{128}}\cdot(x-2)^4

Simplify a few terms using exponent rules:

\sqrt{2}+\frac{x-2}{2\sqrt{2}}-\frac{(x-2)^2}{16\sqrt{2}}+\frac{(x-2)^3}{64\sqrt{2}} -\frac{5(x-2)^4}{1024\sqrt{2}}

This is the fourth-degree Taylor polynomial centered at $x=2$ for $\sqrt{x}$ .

Common Misconceptions

A Taylor polynomial equals the function everywhere. It only approximates $f$ , and it is most accurate near the center $x=a$ . Accuracy usually gets worse as $x$ moves away from $a$ .
Forgetting the factorial. The coefficient is $\frac{f^{(n)}(a)}{n!}$ , not just $f^{(n)}(a)$ . Skipping the $n!$ is one of the most common errors.
Mixing up center and degree. A "third-degree" polynomial stops at the $(x-a)^3$ term. The degree tells you how far to go, not where the center is.
Dropping the $(x-a)$ form when the center is not zero. For a Taylor polynomial about $x=1$ , keep the powers as $(x-1)^n$ . Only Maclaurin polynomials use plain powers of $x$ .
Assuming more terms always help equally. Higher degree usually improves the fit, but only over the interval where the series converges. Far outside that range, adding terms may not help.
Evaluating derivatives at $x$ instead of $a$ . You must plug in the center $a$ for every derivative before building the term.

Vocabulary

The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.

Term	Definition
centered at	The point around which a Taylor polynomial is constructed; the point where the polynomial and function share the same value and derivatives.
coefficient	The numerical factor in front of a term in a polynomial, calculated as the nth derivative of the function divided by n factorial.
degree	The highest power of the variable in a polynomial term.
derivative	The instantaneous rate of change of a function at a specific point, representing the slope of the tangent line to the function at that point.
factorial	The product of all positive integers up to a given number, denoted as n!, used in the denominator of Taylor polynomial coefficients.
function approximation	Using a simpler function, such as a polynomial, to estimate the values of a more complex function.
interval	A connected set of real numbers, typically expressed as a range between two endpoints.
Taylor polynomial	A finite polynomial that approximates a function, formed by taking a partial sum of the Taylor series for that function.

Frequently Asked Questions

How do Taylor polynomial approximations work?

A Taylor polynomial approximates a function near a center x equals a by matching the function and its derivatives at that center. The nth-degree term uses the coefficient f superscript n of a divided by n factorial, multiplied by x minus a to the nth power.

What is the Taylor polynomial coefficient formula?

The coefficient of the nth-degree term in a Taylor polynomial for f centered at x equals a is f superscript n of a divided by n factorial. The power attached to that coefficient is x minus a to the n.

What is a Maclaurin polynomial?

A Maclaurin polynomial is a Taylor polynomial centered at x equals 0. It uses the same Taylor formula, but every derivative is evaluated at 0 and the powers are plain powers of x.

How do you approximate a function value with a Taylor polynomial?

Build the Taylor polynomial to the requested degree, then plug the input value into the polynomial. The approximation is most reliable when the input is close to the center of the polynomial.

What is the difference between a Taylor polynomial and Taylor series?

A Taylor polynomial is a finite partial sum with a set degree. A Taylor series is the full infinite series. Topic 10.11 focuses on building and using the finite polynomial approximation.

How is Taylor polynomial approximation tested on AP Calculus BC?

AP Calculus BC may ask you to build a Taylor or Maclaurin polynomial, use it to approximate a nearby value, or identify a coefficient. Show the center, derivative values, factorials, and powers of x minus a clearly.