Linear approximations | Contextual uses | Achievable AP Calculus AB

What you’ll learn:

Use the tangent line equation to approximate function values near a known point
Understand how concavity affects whether approximations over- or underestimate actual values

When you don’t know the exact value of a function, you can still get a quick estimate with a tangent line. Since the tangent line closely resembles a function near the point of tangency, its equation can be used to approximate values near the known point. This is a technique known as local linear approximation (or tangent line approximation).

Linearization $L (x)$

$L (x) = f (a) + f^{'} (a) (x - a)$

Notice that this is just the formula for the equation of the tangent line at $x = a$ . By plugging in some value $x_{1}$ close to $a,$ the linear approximation $L (x_{1})$ estimates the actual value that is $f (x_{1})$ .

Example

1. Estimate $2.0 1^{3}$ using a linear approximation.

Solution

1. Determine $f (x)$ and the known point.

Define

$f (x) = x^{3}$

$x = 2.01$ is close to $a = 2$ and $f (2) = 8$ is an easy calculation.

2. Find the equation of the tangent line at the known point/point of tangency.

The derivative of $f (x)$ is

$f^{'} (x) = 3 x^{2}$

At the point of tangency $(2, 8)$ , the slope of the tangent line is

$f^{'} (2) = 3 (2)^{2} = 12$

In point-slope form, the equation is

$y - 8 = 12 (x - 2)$

Moving the known $y$ -coordinate to the other side,

$y = 8 + 12 (x - 2)$

Replace $y$ with $L (x)$ . This equation is the linearization of $f (x)$ at $x = 2$ .

$L (x) = 8 + 12 (x - 2)$

3. Approximate the desired function value

We want the function value for $x = 2.01$ , so plug that into $x$ :

$L (2.01) = 8 + 12 (2.01 - 2)$

$= 8 + 12 (0.01)$

$= 8.12$

The actual value (using a calculator) is $8.120601$ , so our estimate is pretty close!

2. Estimate $15$ with a linear approximation.

Solution

(spoiler)

1. Determine $f (x)$ and the known point.

$f (x) = x$

$x = 15$ is close to $a = 16$ and $f (16) = 4$ . So the known point is $(16, 4)$ . For now, we at least know that $15$ will be a little less than $4$ .

2. Find the equation of the tangent line at the known point/point of tangency.

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

$f^{'} (16) = \frac{1}{2 16}$

$= \frac{1}{8}$

Then the linear approximation formula is:

$L (x) = 4 + \frac{1}{8} (x - 16)$

3. Approximate the desired function value

Plug in $15$ for $x$ .

$L (15) = 4 + \frac{1}{8} (15 - 16)$

$= 3 \frac{7}{8}$

$= 3.875$

Inputting $15$ into a calculator gives $\approx 3.872 \dots$ so our estimate is fairly close!

Curvature

Linear approximations are more accurate the closer $x$ is to $a$ , but the accuracy also depends on how curved the function is near that point.

The 2nd derivative $f^{''} (x)$ tells us about the concavity of a function - whether the graph bends upward or downward. More on concavity and its uses will be discussed later. For now, just remember:

If $f^{''} (x) > 0$ , then $f (x)$ is concave up - shaped “like a cup.”
- Tangent lines will lie below $f (x)$ and underestimate the actual value.
If $f^{''} (x) < 0$ , then $f (x)$ is concave down - shaped “like a frown.”
- Tangent lines will lie above $f (x)$ and overestimate the actual value.

The following image shows how tangent lines always lie below a curve that is concave up and above a curve that is concave down.

Figure 4.4.2 Tangent lines and concavity

3. Approximate $ln (1.1)$ . Is this an overestimate or underestimate?

Solution

(spoiler)

1. Determine $f (x)$ and the known point.

$f (x) = ln (x)$ .

The known point is $(1, 0)$ since $ln (1) = 0$ .

2. Find the equation of the tangent line at the known point/point of tangency.

$f^{'} (x) = \frac{1}{x}$

$f^{'} (1) = 1$

Then the equation of the tangent line is

$y - 0 = 1 (x - 1)$

The linearization of $f (x)$ at $x = 1$ is

$L (x) = x - 1$

3. Approximate the desired function value

The approximation of $ln (1.1)$ is

$L (1.1) = 1.1 - 1$

$= 0.1$

Without using a calculator, we can determine if this under- or over-estimates the actual value. The 2nd derivative is

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

$f^{''} (x) < 0$ for any value of $x$ , which means the graph of $f (x)$ is entirely concave down. So the tangent line approximation is an *overestimate." This can be confirmed by entering $ln (1.1)$ into a calculator, which gives $\approx 0.095 \dots$ .

Estimate $e^{0.02}$ with a linear approximation. Is this an overestimate or an underestimate?

Solution

(spoiler)

Let’s define $f (x) = e^{x}$ .

1. Determine $f (x)$ and the known point.

$x = 0.02$ is close to $a = 0$ . So the known point is $(0, 1)$ .

2. Find the equation of the tangent line at the known point/point of tangency.

$f^{'} (x) = e^{x}$

$f^{'} (0) = 1$

Then the linearization at $x = 0$ is:

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

$= 1 + x$

3. Approximate the desired function value

Plug in $x = 0.02$ to estimate $e^{0.02}$ :

$L (0.02) = 1 + 0.02$

$= 1.02$

To determine if this is an overestimate or an underestimate, determine the concavity at $a = 0$ :

$f^{''} (x) = e^{x}$

$f^{''} (0) = 1$

Even without plugging in $a$ , we know that $e^{x}$ is always positive no matter what value is inputted. So $f (x) = e^{x}$ is always concave up and all linear approximations will underestimate the actual value. The calculator confirms that $e^{0.02} \approx 1.0202 \dots$ which is greater than just $1.02$ .

What you’ll learn:

Use the tangent line equation to approximate function values near a known point
Understand how concavity affects whether approximations over- or underestimate actual values

Linearization $L (x)$

$L (x) = f (a) + f^{'} (a) (x - a)$

Example

1. Estimate $2.0 1^{3}$ using a linear approximation.

Solution

1. Determine $f (x)$ and the known point.

Define

$f (x) = x^{3}$

$x = 2.01$ is close to $a = 2$ and $f (2) = 8$ is an easy calculation.

2. Find the equation of the tangent line at the known point/point of tangency.

The derivative of $f (x)$ is

$f^{'} (x) = 3 x^{2}$

At the point of tangency $(2, 8)$ , the slope of the tangent line is

$f^{'} (2) = 3 (2)^{2} = 12$

In point-slope form, the equation is

$y - 8 = 12 (x - 2)$

Moving the known $y$ -coordinate to the other side,

$y = 8 + 12 (x - 2)$

Replace $y$ with $L (x)$ . This equation is the linearization of $f (x)$ at $x = 2$ .

$L (x) = 8 + 12 (x - 2)$

3. Approximate the desired function value

We want the function value for $x = 2.01$ , so plug that into $x$ :

$L (2.01) = 8 + 12 (2.01 - 2)$

$= 8 + 12 (0.01)$

$= 8.12$

The actual value (using a calculator) is $8.120601$ , so our estimate is pretty close!

2. Estimate $15$ with a linear approximation.

Solution

(spoiler)

1. Determine $f (x)$ and the known point.

$f (x) = x$

$x = 15$ is close to $a = 16$ and $f (16) = 4$ . So the known point is $(16, 4)$ . For now, we at least know that $15$ will be a little less than $4$ .

2. Find the equation of the tangent line at the known point/point of tangency.

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

$f^{'} (16) = \frac{1}{2 16}$

$= \frac{1}{8}$

Then the linear approximation formula is:

$L (x) = 4 + \frac{1}{8} (x - 16)$

3. Approximate the desired function value

Plug in $15$ for $x$ .

$L (15) = 4 + \frac{1}{8} (15 - 16)$

$= 3 \frac{7}{8}$

$= 3.875$

Inputting $15$ into a calculator gives $\approx 3.872 \dots$ so our estimate is fairly close!

Curvature

Linear approximations are more accurate the closer $x$ is to $a$ , but the accuracy also depends on how curved the function is near that point.

If $f^{''} (x) > 0$ , then $f (x)$ is concave up - shaped “like a cup.”
- Tangent lines will lie below $f (x)$ and underestimate the actual value.
If $f^{''} (x) < 0$ , then $f (x)$ is concave down - shaped “like a frown.”
- Tangent lines will lie above $f (x)$ and overestimate the actual value.

The following image shows how tangent lines always lie below a curve that is concave up and above a curve that is concave down.

3. Approximate $ln (1.1)$ . Is this an overestimate or underestimate?

Solution

(spoiler)

1. Determine $f (x)$ and the known point.

$f (x) = ln (x)$ .

The known point is $(1, 0)$ since $ln (1) = 0$ .

2. Find the equation of the tangent line at the known point/point of tangency.

$f^{'} (x) = \frac{1}{x}$

$f^{'} (1) = 1$

Then the equation of the tangent line is

$y - 0 = 1 (x - 1)$

The linearization of $f (x)$ at $x = 1$ is

$L (x) = x - 1$

3. Approximate the desired function value

The approximation of $ln (1.1)$ is

$L (1.1) = 1.1 - 1$

$= 0.1$

Without using a calculator, we can determine if this under- or over-estimates the actual value. The 2nd derivative is

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

Estimate $e^{0.02}$ with a linear approximation. Is this an overestimate or an underestimate?

Solution

(spoiler)

Let’s define $f (x) = e^{x}$ .

1. Determine $f (x)$ and the known point.

$x = 0.02$ is close to $a = 0$ . So the known point is $(0, 1)$ .

2. Find the equation of the tangent line at the known point/point of tangency.

$f^{'} (x) = e^{x}$

$f^{'} (0) = 1$

Then the linearization at $x = 0$ is:

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

$= 1 + x$

3. Approximate the desired function value

Plug in $x = 0.02$ to estimate $e^{0.02}$ :

$L (0.02) = 1 + 0.02$

$= 1.02$

To determine if this is an overestimate or an underestimate, determine the concavity at $a = 0$ :

$f^{''} (x) = e^{x}$

$f^{''} (0) = 1$