AP Calculus AB

## Learning Outcomes

- Write the tangent line to $$y=f(x)$$ at $$x=a$$ and use it to approximate $$f(x)$$ near $$a$$.
- Interpret $$f'(a)(x-a)$$ as an approximate change in the value of a function.
- Use local linearization to estimate values of trigonometric, polynomial, rational, radical, and implicitly defined functions.
- Apply the common approximation $$(1+x)^k\approx 1+kx$$ for $$x$$ near $$0$$.
- Decide whether a tangent-line approximation is an overestimate or underestimate using concavity.
- Solve AP-style contextual approximation problems involving volume, area, and measurement error.

## AP Calculus AB Syllabus

In AP Calculus AB, tangent line approximation appears in Unit 4 as an application of derivatives. The derivative gives the slope of the tangent line, and that tangent line can be used as a local linear model for the function near the point of tangency. On the exam, this skill may appear as a direct approximation problem, a contextual “approximate change” problem, or part of a free-response question involving a graph, table, differential equation, or implicit relation.

- The tangent line to $$y=f(x)$$ at $$x=a$$ has slope $$f'(a)$$ and passes through $$(a,f(a))$$.
- The local linear approximation is $$f(x)\approx f(a)+f'(a)(x-a)$$ for $$x$$ close to $$a$$.
- The approximate change in the function is $$\Delta f\approx f'(a)\Delta x$$.
- A tangent-line approximation is usually more accurate when $$x$$ is closer to $$a$$.
- Concavity determines whether the tangent line lies above or below the graph near the point.
- Contextual problems may ask for an approximate value, an approximate increase or decrease, or a possible measurement error.
- For implicitly defined functions, first find $$\frac{dy}{dx}$$, then use the tangent line at the given point.

## Test Your Knowledge

Attempt these questions before reading this article. If you find some difficult or cannot remember the answers, look more closely at that area during your revision.

1. If $$f(2)=5$$ and $$f'(2)=-3$$, what is the tangent-line approximation for $$f(2.1)$$?
2. Write the linearization of $$f(x)=\sqrt{1+x}$$ at $$x=0$$.
3. Why does a tangent line to a concave-up graph give an underestimate near the point of tangency?
4. The volume of a sphere is $$V=\frac43\pi r^3$$. Use a tangent line to approximate the increase in volume when $$r$$ increases from $$3$$ to $$3.1$$.
5. For the implicit equation $$2x^2-y^3=10$$, given that $$(3,2)$$ lies on the graph, how would you approximate $$y$$ when $$x=3.04$$?

## Introduction

A tangent line gives more than just the slope of a curve at one point. Near the point of tangency, the graph of a differentiable function often looks very much like its tangent line. AP Calculus AB uses this idea to estimate function values without fully evaluating the original function.

For example, near $$x=0$$, the graph of $$y=\sin x$$ is close to its tangent line $$y=x$$. That is why $$\sin(0.05)\approx 0.05$$. The estimate is not exact, but for values close to the point of tangency it can be very accurate.

> **Key Term: tangent-line approximation**
> A method of estimating $$f(x)$$ by using the $$y$$-value on the tangent line to $$y=f(x)$$ at a nearby point $$x=a$$.

### Building the Tangent Line Approximation Formula

Suppose $$f$$ is differentiable at $$x=a$$. The tangent line to $$y=f(x)$$ at $$x=a$$ passes through the point $$(a,f(a))$$ and has slope $$f'(a)$$. Using point-slope form, its equation is

y-f(a)=f'(a)(x-a)
Solving for $$y$$ gives

y=f(a)+f'(a)(x-a)
The central idea of tangent-line approximation is that, for $$x$$ close to $$a$$,

f(x)\approx f(a)+f'(a)(x-a)
The expression on the right is often called the linear approximation or linearization of $$f$$ at $$a$$.

> **Key Term: local linear approximation**
> An approximation of a function near a point using the tangent line at that point.
>
> **Key Term: linearization**
> The linear function $$L(x)=f(a)+f'(a)(x-a)$$ used to approximate $$f(x)$$ near $$x=a$$.

The word “local” matters. A tangent line is usually a good model only close to the point of tangency. The farther $$x$$ moves away from $$a$$, the more the graph may bend away from the line.

> **Test Tip:**
> In an AP problem, identify three pieces immediately: the base point $$a$$, the known function value $$f(a)$$, and the derivative value $$f'(a)$$. Then substitute into $$L(x)=f(a)+f'(a)(x-a)$$.

Consider a simple data-style example.

> **Question:**
> Suppose $$f(4)=10$$ and $$f'(4)=1.5$$. Approximate $$f(4.2)$$.

Here $$a=4$$, $$f(a)=10$$, $$f'(a)=1.5$$, and $$x=4.2$$.

$$
f(4.2)\approx 10+1.5(4.2-4)
f(4.2)\approx 10+1.5(0.2)=10.3
$$

So the tangent-line approximation gives

f(4.2)\approx 10.3
This format is common on AP free-response questions because it tests whether you can use derivative information to estimate a nearby function value.

### Finding Linearizations from Formulas

When a function is given by a formula, you usually find the linearization by differentiating first, then evaluating at the point of tangency.

The general procedure is:

1. Choose the point $$a$$.
2. Compute $$f(a)$$.
3. Compute $$f'(x)$$.
4. Compute $$f'(a)$$.
5. Write $$L(x)=f(a)+f'(a)(x-a)$$.
6. Substitute the target $$x$$ if an approximation is requested.

> **Question:**
> Find the tangent-line approximation for $$f(x)=2x^3-3x$$ at $$a=1$$.

First evaluate the function:

$$
f(1)=2(1)^3-3(1)=-1
$$

Differentiate:

f'(x)=6x^2-3
Then

f'(1)=6(1)^2-3=3
So the linearization is

$$
L(x)=-1+3(x-1)
$$

or

$$
L(x)=3x-4
$$

Therefore, near $$x=1$$,

$$
2x^3-3x\approx 3x-4
$$

For example,

$$
f(1.1)\approx 3(1.1)-4=-0.7
$$

The true value is

$$
2(1.1)^3-3(1.1)=2(1.331)-3.3=-0.638
$$

The approximation is close, though not exact.

Now compare this with a trigonometric example.

> **Question:**
> Find the tangent-line approximation for $$\sin x$$ at $$a=0$$.

Here

$$
f(x)=\sin x
$$

so

$$
f(0)=0
$$

and

f'(x)=\cos x
Thus

f'(0)=1
The linearization is

$$
L(x)=0+1(x-0)=x
$$

So, for $$x$$ near $$0$$,

\sin x\approx x
This is a major small-angle approximation. In AP Calculus AB, radians are assumed for trigonometric calculus. The approximation $$\sin x\approx x$$ works near $$0$$ only when $$x$$ is measured in radians.

> **Exam Warning:**
> Do not use degree measure in trigonometric linearization problems unless a problem specifically tells you to convert. Calculus formulas such as $$\frac{d}{dx}\sin x=\cos x$$ assume radians.

A second common trigonometric example is $$\cos x$$ near $$a=\frac{\pi}{2}$$.

$$
f(x)=\cos x
f\left(\frac{\pi}{2}\right)=0
$$

f'(x)=-\sin x
f'\left(\frac{\pi}{2}\right)=-1
So

$$
L(x)=0-1\left(x-\frac{\pi}{2}\right)
$$

which simplifies to

$$
L(x)=-x+\frac{\pi}{2}
$$

Thus, for $$x$$ near $$\frac{\pi}{2}$$,

$$
\cos x\approx -x+\frac{\pi}{2}
$$

### Using Approximate Change in Context

Tangent-line approximation is also written in change form. If $$x$$ changes from $$a$$ to $$a+\Delta x$$, then

\Delta f\approx f'(a)\Delta x
This says that the change in the output is approximately the derivative times the change in the input.

> **Key Term: differential**
> A small change approximation written as $$dy=f'(x)\,dx$$, often used to estimate the change in $$y$$ caused by a small change $$dx$$ in $$x$$.

On the AP exam, the language may be “approximate the increase,” “approximate the change,” or “estimate the error.” These usually point to

\Delta f\approx f'(a)\Delta x
not necessarily to the full value of $$f(x)$$.

> **Question:**
> The volume of a sphere is $$V=\frac43\pi r^3$$. Use a tangent line to approximate the increase in volume when the radius increases from $$3$$ inches to $$3.1$$ inches.

Here the function is volume as a function of radius:

$$
V(r)=\frac43\pi r^3
$$

The radius changes from $$3$$ to $$3.1$$, so

$$
\Delta r=0.1
$$

Differentiate:

V'(r)=4\pi r^2
At $$r=3$$,

V'(3)=4\pi(3)^2=36\pi
Therefore the approximate change in volume is

\Delta V\approx V'(3)\Delta r

$$
\Delta V\approx 36\pi(0.1)=3.6\pi
$$

So the increase in volume is approximately

3.6\pi
cubic inches.

Notice that the question asks for the increase, not the new volume. The original volume at $$r=3$$ is

$$
V(3)=36\pi
$$

The approximate new volume would be

$$
36\pi+3.6\pi=39.6\pi
$$

But the approximate increase is only $$3.6\pi$$.

> **Exam Warning:**
> Read whether the question asks for the approximate function value or the approximate change. $$f(a)+f'(a)\Delta x$$ gives the new value, while $$f'(a)\Delta x$$ gives only the change.

Implicit relations can also use tangent-line approximation. First find the derivative by implicit differentiation.

> **Question:**
> When $$x=3$$, the equation $$2x^2-y^3=10$$ has the solution $$y=2$$. Use a tangent line to approximate $$y$$ when $$x=3.04$$.

Differentiate both sides with respect to $$x$$:

$$
4x-3y^2\frac{dy}{dx}=0
$$

Solve for $$\frac{dy}{dx}$$:

$$
\frac{dy}{dx}=\frac{4x}{3y^2}
$$

At the point $$(3,2)$$,

$$
\frac{dy}{dx}=\frac{4(3)}{3(2)^2}=\frac{12}{12}=1
$$

The tangent line at $$(3,2)$$ is

$$
y=2+1(x-3)
$$

At $$x=3.04$$,

$$
y\approx 2+(3.04-3)=2.04
$$

So the estimate is

y\approx 2.04

### The Special Approximation $$(1+x)^k\approx 1+kx$$

A very useful tangent-line approximation comes from the function

$$
f(x)=(1+x)^k
$$

where $$k$$ is any real number and $$x$$ is near $$0$$.

At $$x=0$$,

$$
f(0)=(1+0)^k=1
$$

Differentiate:

f'(x)=k(1+x)^{k-1}
Then

f'(0)=k(1+0)^{k-1}=k
So the linearization at $$0$$ is

$$
L(x)=1+kx
$$

Therefore,

$$
(1+x)^k\approx 1+kx
$$

for $$x$$ near $$0$$.

For example, if $$k=\frac12$$, then

$$
\sqrt{1+x}=(1+x)^{1/2}\approx 1+\frac12x
$$

For $$x=0.04$$,

$$
\sqrt{1.04}\approx 1+\frac12(0.04)=1.02
$$

This is close to the true value.

If $$k=-2$$, then

$$
(1+x)^{-2}\approx 1-2x
$$

This kind of approximation can turn a difficult expression into a quick mental estimate.

Sometimes you must rewrite the expression so that it has the form $$(1+x)^k$$ near $$x=0$$.

> **Question:**
> Estimate $$\frac{3}{(1-x)^2}$$ at $$x=0.05$$ using a tangent line at $$x=0$$.

Let

$$
f(x)=\frac{3}{(1-x)^2}=3(1-x)^{-2}
$$

At $$x=0$$,

$$
f(0)=3
$$

Differentiate:

f'(x)=6(1-x)^{-3}=\frac{6}{(1-x)^3}
So

f'(0)=6
The tangent line is

$$
L(x)=3+6x
$$

At $$x=0.05$$,

$$
f(0.05)\approx 3+6(0.05)=3.3
$$

So the estimate is

3.3

### Overestimates, Underestimates, and Concavity

AP questions often ask whether a tangent-line approximation is an overestimate or an underestimate. This is controlled by concavity.

> **Key Term: concavity**
> The direction a graph bends. If $$f''(x)>0$$, the graph is concave up; if $$f''(x)<0$$, the graph is concave down.

For a concave-up graph, the tangent line lies below the graph near the point of tangency. Therefore, the tangent-line approximation is an underestimate.

For a concave-down graph, the tangent line lies above the graph near the point of tangency. Therefore, the tangent-line approximation is an overestimate.

> **Test Tip:**
> To justify overestimate or underestimate on a free-response question, connect the sign of $$f''$$ to concavity, then connect concavity to the tangent line’s position relative to the graph.

Return to the previous example:

$$
f(x)=\frac{3}{(1-x)^2}
$$

We found

f'(x)=\frac{6}{(1-x)^3}
Differentiate again:

f''(x)=\frac{18}{(1-x)^4}
For $$x\ne 1$$,

f''(x)>0
Thus $$f$$ is concave up on intervals where it is defined. On the interval from $$0$$ to $$0.05$$, the graph is concave up, so the tangent line at $$x=0$$ lies below the graph. Therefore, the estimate

f(0.05)\approx 3.3
is an underestimate.

Be careful: “increasing” and “concave up” are not the same idea. Increasing means $$f'(x)>0$$. Concave up means $$f''(x)>0$$. The overestimate or underestimate decision depends on concavity, not simply on whether the function is increasing or decreasing.

Here is another example.

> **Question:**
> Use a tangent line to approximate $$\sqrt{1+x}$$ at $$x=0.1$$. Is the estimate an overestimate or underestimate?

Let

$$
f(x)=\sqrt{1+x}
$$

At $$a=0$$,

$$
f(0)=1
$$

and

f'(x)=\frac{1}{2\sqrt{1+x}}
so

f'(0)=\frac12
The linearization is

$$
L(x)=1+\frac12x
$$

At $$x=0.1$$,

$$
\sqrt{1.1}\approx 1+\frac12(0.1)=1.05
$$

Now check concavity:

f''(x)=-\frac{1}{4}(1+x)^{-3/2}
For $$x>-1$$,

f''(x)<0
So the graph is concave down. The tangent line lies above the graph, and the estimate $$1.05$$ is an overestimate.

## Key Point Checklist

This article has covered the following key knowledge points:

- The tangent line to $$y=f(x)$$ at $$x=a$$ is $$y-f(a)=f'(a)(x-a)$$.
- The local linear approximation is $$f(x)\approx f(a)+f'(a)(x-a)$$.
- The linearization function is usually written $$L(x)=f(a)+f'(a)(x-a)$$.
- The approximation is generally better when $$x$$ is close to $$a$$.
- The term $$f'(a)(x-a)$$ represents the approximate change in $$f$$.
- In change notation, $$\Delta f\approx f'(a)\Delta x$$.
- Contextual problems may ask for either the new approximate value or only the approximate change.
- For implicit curves, find $$\frac{dy}{dx}$$ first, evaluate the slope at the given point, then write the tangent line.
- The approximation $$(1+x)^k\approx 1+kx$$ works for $$x$$ near $$0$$.
- If $$f''>0$$, the graph is concave up and the tangent-line estimate is an underestimate.
- If $$f''<0$$, the graph is concave down and the tangent-line estimate is an overestimate.
- Trigonometric tangent-line approximations in calculus use radian measure.

## Key Terms and Concepts

- tangent-line approximation
- local linear approximation
- linearization
- differential
- concavity


Approximating values of a function using local linearity and linearization - Tangent line approximation