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Reading Time: 4 min
Last Updated: March 30, 2026
Main Ideas: 4
Reading Time: 4 min
Last Updated: March 30, 2026
Main Ideas: 4

Topic 10.11 Notes – Finding Taylor Polynomial Approximations of Functions

Verified for 2027 AP® Calculus BC Exam
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These are called Taylor polynomials. You’ll learn how to construct them and how to use them to approximate function values close to the center.

What a Taylor polynomial is

A Taylor polynomial is a polynomial that matches a function’s value and derivatives at a specific point x=ax = a.

It’s built from the derivatives of ff evaluated at aa.

The general form is:

Tn(x)=∑k=0nf(k)(a)k!(x−a)k T_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k

Written out:

f(a)+f′(a)(x−a)+f′′(a)2!(x−a)2+⋯+f(n)(a)n!(x−a)n f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \dots + \frac{f^{(n)}(a)}{n!}(x-a)^n

Lock these in:

  • The coefficient of (x−a)n(x-a)^n is f(n)(a)n! \frac{f^{(n)}(a)}{n!} .
  • f(0)(a)=f(a)f^{(0)}(a) = f(a).
  • A Maclaurin polynomial is just centered at a=0a=0.
  • Higher degree → matches more derivatives → usually better approximation near aa.

Think of it as forcing a polynomial to have the same value, slope, concavity, and higher-order behavior as the function at one point.

Here’s the visual idea:

Maclaurin polynomials for sin⁡x \sin x (degrees 1, 3, and 5)

Notice how they agree closely near 0 but drift away farther out. Taylor polynomials are local approximations.

How to build a Taylor polynomial

When you’re asked for, say, the 3rd-degree Taylor polynomial centered at x=ax=a, the structure is always the same.

Step-by-step

  1. Compute derivatives up to the 3rd derivative.
  2. Evaluate each at x=ax=a.
  3. Plug into
    f(k)(a)k!(x−a)k \frac{f^{(k)}(a)}{k!}(x-a)^k
  4. Add terms from k=0k=0 to k=3k=3.

Quick example

Find the second-degree Taylor polynomial for f(x)=ln⁡xf(x)=\ln x centered at a=2a=2.

Derivatives:

  • f(x)=ln⁡xf(x)=\ln x
  • f′(x)=1/xf'(x)=1/x
  • f′′(x)=−1/x2f''(x)=-1/x^2

Evaluate at 2:

  • f(2)=ln⁡2f(2)=\ln 2
  • f′(2)=1/2f'(2)=1/2
  • f′′(2)=−1/4f''(2)=-1/4

Build the polynomial:

T2(x)=ln⁡2+12(x−2)+−1/42(x−2)2 T_2(x) = \ln 2 + \frac{1}{2}(x-2) + \frac{-1/4}{2}(x-2)^2

T2(x)=ln⁡2+12(x−2)−18(x−2)2 T_2(x) = \ln 2 + \frac{1}{2}(x-2) - \frac{1}{8}(x-2)^2

Every coefficient came directly from f(k)(a)k! \frac{f^{(k)}(a)}{k!} . That’s the core skill.

On tests, most errors happen from:

  • Forgetting the factorial
  • Using xnx^n instead of (x−a)n(x-a)^n
  • Stopping at the wrong degree

What improves as degree increases

Each added term forces another derivative to match at x=ax=a.

So as nn increases:

  • The graphs agree more tightly near aa.
  • The polynomial often stays accurate over a slightly larger interval.
  • The approximation error typically shrinks near the center.

But this is always near aa. Far away, even high-degree polynomials can behave wildly.

That idea connects to the larger theme of Unit 10: power series represent functions over certain intervals. A Taylor polynomial is just a finite “snapshot” of that full series.

Using a Taylor polynomial to approximate values

Taylor polynomials are used to approximate f(b)f(b) when bb is close to aa.

Example idea: suppose you built a third-degree Maclaurin polynomial for exe^x:

T3(x)=1+x+x22+x36 T_3(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}

To approximate e0.2e^{0.2}, plug in x=0.2x=0.2. That’s it.

Key idea:

  • The closer bb is to aa, the better the approximation.
  • Higher degree usually improves accuracy.

On the AP exam, this often shows up as:

  • “Use the third-degree Taylor polynomial to approximate…”
  • Calculator section problems where you evaluate your polynomial numerically.
  • FRQs where you must actually write the polynomial first, then substitute.

Key Takeaways

The coefficient of (x−a)n(x-a)^n is always f(n)(a)n! \frac{f^{(n)}(a)}{n!} .
A Maclaurin polynomial is just a Taylor polynomial with a=0a=0.
An nth-degree Taylor polynomial must stop at power nn.
Taylor polynomials approximate functions best near the center aa.
If your polynomial doesn’t give f(a)f(a) when you plug in x=ax=a, something is wrong.

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Notes

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