4.4 Linear approximations

Achievable AP Calculus AB

4. Contextual uses

Our AP Calculus AB course is currently in development and is a work-in-progress.

Linear approximations

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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 often get a quick estimate using the tangent line. Near the point of tangency, the tangent line closely matches the function, so its equation gives a good approximation for nearby inputs. This method is called local linear approximation (also called tangent line approximation).

Linearization $L (x)$

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

This is the tangent line to $f (x)$ at $x = a$ . If $x_{1}$ is close to $a$ , then $L (x_{1})$ is an estimate of the actual value $f (x_{1})$ .

Example

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

Solution

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

Define

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

Since $2.01$ is close to $2$ , choose $a = 2$ . Then

$f (2) = 8$

so the known point is $(2, 8)$ .

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

Differentiate:

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

Evaluate the slope at $x = 2$ :

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

Use point-slope form at $(2, 8)$ :

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

Solve for $y$ :

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

Replace $y$ with $L (x)$ to write the linearization at $x = 2$ :

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

3. Approximate the desired function value

Plug in $x = 2.01$ :

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

$= 8 + 12 (0.01)$

$= 8.12$

The actual value (using a calculator) is $8.120601$ , so the linear approximation is very close.

Estimate $15$ with a linear approximation.

Solution

(spoiler)

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

Let

$f (x) = x$

Since $15$ is close to $16$ and $16$ is easy to compute, choose $a = 16$ . Then

$f (16) = 4$

so the known point is $(16, 4)$ . This also tells us $15$ should be a little less than $4$ .

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

Differentiate:

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

Evaluate at $x = 16$ :

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

$= \frac{1}{8}$

Use the linearization formula:

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

3. Approximate the desired function value

Plug in $x = 15$ :

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

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

$= 3.875$

A calculator gives $15 \approx 3.872 \dots$ , so the estimate is close.

Curvature

Linear approximations get better as $x$ gets closer to $a$ . Accuracy also depends on how much the function curves near $x = a$ .

The second derivative $f^{''} (x)$ describes the concavity of a function - whether the graph bends upward or downward. More on concavity and its uses will come later. For now, keep these ideas in mind:

If $f^{''} (x) > 0$ , then $f (x)$ is concave up (shaped “like a cup”).
- The tangent line lies below $f (x)$ , so the linear approximation will underestimate the actual value.
If $f^{''} (x) < 0$ , then $f (x)$ is concave down (shaped “like a frown”).
- The tangent line lies above $f (x)$ , so the linear approximation will overestimate the actual value.

The image below shows this relationship between tangent lines and concavity.

Figure 4.4.2 Tangent lines and concavity

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

Solution

(spoiler)

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

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

A convenient known point is $x = 1$ because

$ln (1) = 0$

So the known point is $(1, 0)$ .

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

Differentiate:

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

Evaluate at $x = 1$ :

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

Write the tangent line at $(1, 0)$ :

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

So the linearization at $x = 1$ is

$L (x) = x - 1$

3. Approximate the desired function value

Plug in $x = 1.1$ :

$L (1.1) = 1.1 - 1$

$= 0.1$

To decide whether this is an overestimate or underestimate, look at concavity:

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

Since $f^{''} (x) < 0$ for all $x > 0$ , $ln (x)$ is concave down everywhere in its domain. That means the tangent line lies above the curve, so the tangent line approximation is an overestimate.

A calculator confirms that $ln (1.1) \approx 0.095 \dots$ , which is slightly less than $0.1$ .

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

Solution

(spoiler)

Let $f (x) = e^{x}$ .

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

Since $0.02$ is close to $0$ , choose $a = 0$ . Then

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

so the known point is $(0, 1)$ .

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

Differentiate:

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

Evaluate at $x = 0$ :

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

Write the linearization at $x = 0$ :

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

$= 1 + x$

3. Approximate the desired function value

Plug in $x = 0.02$ :

$L (0.02) = 1 + 0.02$

$= 1.02$

To decide whether this is an overestimate or underestimate, check concavity:

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

Because $e^{x} > 0$ for all $x$ , $f (x) = e^{x}$ is always concave up. That means the tangent line lies below the curve, so the linear approximation is an underestimate.

A calculator confirms that $e^{0.02} \approx 1.0202 \dots$ , which is slightly greater than $1.02$ .

Local linear approximation (tangent line approximation)

Uses tangent line at $x = a$ to estimate $f (x)$ near $a$
Linearization formula: $L (x) = f (a) + f^{'} (a) (x - a)$
Good for quick estimates when $x$ is close to $a$

Example: Estimating $2.0 1^{3}$

Choose $f (x) = x^{3}$ , $a = 2$
$f (2) = 8$ , $f^{'} (2) = 12$
Linearization: $L (x) = 8 + 12 (x - 2)$
$L (2.01) = 8.12$ (close to actual $8.120601$ )

Example: Estimating $15$

Choose $f (x) = x$ , $a = 16$
$f (16) = 4$ , $f^{'} (16) = \frac{1}{8}$
Linearization: $L (x) = 4 + \frac{1}{8} (x - 16)$
$L (15) = 3.875$ (close to actual $3.872$ )

Curvature and concavity

Accuracy improves as $x$ approaches $a$
Concavity determined by $f^{''} (x)$ :
- $f^{''} (x) > 0$ : concave up, tangent line underestimates
- $f^{''} (x) < 0$ : concave down, tangent line overestimates

Example: Estimating $ln (1.1)$

$f (x) = ln (x)$ , $a = 1$
$f (1) = 0$ , $f^{'} (1) = 1$
Linearization: $L (x) = x - 1$
$L (1.1) = 0.1$ (overestimate, since $f^{''} (x) < 0$ )

Example: Estimating $e^{0.02}$

$f (x) = e^{x}$ , $a = 0$
$f (0) = 1$ , $f^{'} (0) = 1$
Linearization: $L (x) = 1 + x$
$L (0.02) = 1.02$ (underestimate, since $f^{''} (x) > 0$ )

Linear approximations

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 $L (x)$

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

This is the tangent line to $f (x)$ at $x = a$ . If $x_{1}$ is close to $a$ , then $L (x_{1})$ is an estimate of the actual value $f (x_{1})$ .

Example

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

Solution

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

Define

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

Since $2.01$ is close to $2$ , choose $a = 2$ . Then

$f (2) = 8$

so the known point is $(2, 8)$ .

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

Differentiate:

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

Evaluate the slope at $x = 2$ :

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

Use point-slope form at $(2, 8)$ :

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

Solve for $y$ :

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

Replace $y$ with $L (x)$ to write the linearization at $x = 2$ :

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

3. Approximate the desired function value

Plug in $x = 2.01$ :

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

$= 8 + 12 (0.01)$

$= 8.12$

The actual value (using a calculator) is $8.120601$ , so the linear approximation is very close.

Estimate $15$ with a linear approximation.

Solution

(spoiler)

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

Let

$f (x) = x$

Since $15$ is close to $16$ and $16$ is easy to compute, choose $a = 16$ . Then

$f (16) = 4$

so the known point is $(16, 4)$ . This also tells us $15$ should be a little less than $4$ .

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

Differentiate:

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

Evaluate at $x = 16$ :

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

$= \frac{1}{8}$

Use the linearization formula:

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

3. Approximate the desired function value

Plug in $x = 15$ :

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

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

$= 3.875$

A calculator gives $15 \approx 3.872 \dots$ , so the estimate is close.

Curvature

Linear approximations get better as $x$ gets closer to $a$ . Accuracy also depends on how much the function curves near $x = a$ .

If $f^{''} (x) > 0$ , then $f (x)$ is concave up (shaped “like a cup”).
- The tangent line lies below $f (x)$ , so the linear approximation will underestimate the actual value.
If $f^{''} (x) < 0$ , then $f (x)$ is concave down (shaped “like a frown”).
- The tangent line lies above $f (x)$ , so the linear approximation will overestimate the actual value.

The image below shows this relationship between tangent lines and concavity.

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

Solution

(spoiler)

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

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

A convenient known point is $x = 1$ because

$ln (1) = 0$

So the known point is $(1, 0)$ .

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

Differentiate:

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

Evaluate at $x = 1$ :

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

Write the tangent line at $(1, 0)$ :

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

So the linearization at $x = 1$ is

$L (x) = x - 1$

3. Approximate the desired function value

Plug in $x = 1.1$ :

$L (1.1) = 1.1 - 1$

$= 0.1$

To decide whether this is an overestimate or underestimate, look at concavity:

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

Since $f^{''} (x) < 0$ for all $x > 0$ , $ln (x)$ is concave down everywhere in its domain. That means the tangent line lies above the curve, so the tangent line approximation is an overestimate.

A calculator confirms that $ln (1.1) \approx 0.095 \dots$ , which is slightly less than $0.1$ .

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

Solution

(spoiler)

Let $f (x) = e^{x}$ .

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

Since $0.02$ is close to $0$ , choose $a = 0$ . Then

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

so the known point is $(0, 1)$ .

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

Differentiate:

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

Evaluate at $x = 0$ :

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

Write the linearization at $x = 0$ :

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

$= 1 + x$

3. Approximate the desired function value

Plug in $x = 0.02$ :

$L (0.02) = 1 + 0.02$

$= 1.02$

To decide whether this is an overestimate or underestimate, check concavity:

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

Because $e^{x} > 0$ for all $x$ , $f (x) = e^{x}$ is always concave up. That means the tangent line lies below the curve, so the linear approximation is an underestimate.

A calculator confirms that $e^{0.02} \approx 1.0202 \dots$ , which is slightly greater than $1.02$ .

Achievable AP Calculus AB

4. Contextual uses

Our AP Calculus AB course is currently in development and is a work-in-progress.

Linear approximations

6 min read

Font

Discuss

Feedback

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 $L (x)$

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

This is the tangent line to $f (x)$ at $x = a$ . If $x_{1}$ is close to $a$ , then $L (x_{1})$ is an estimate of the actual value $f (x_{1})$ .

Example

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

Solution

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

Define

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

Since $2.01$ is close to $2$ , choose $a = 2$ . Then

$f (2) = 8$

so the known point is $(2, 8)$ .

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

Differentiate:

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

Evaluate the slope at $x = 2$ :

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

Use point-slope form at $(2, 8)$ :

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

Solve for $y$ :

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

Replace $y$ with $L (x)$ to write the linearization at $x = 2$ :

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

3. Approximate the desired function value

Plug in $x = 2.01$ :

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

$= 8 + 12 (0.01)$

$= 8.12$

The actual value (using a calculator) is $8.120601$ , so the linear approximation is very close.

Estimate $15$ with a linear approximation.

Solution

(spoiler)

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

Let

$f (x) = x$

Since $15$ is close to $16$ and $16$ is easy to compute, choose $a = 16$ . Then

$f (16) = 4$

so the known point is $(16, 4)$ . This also tells us $15$ should be a little less than $4$ .

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

Differentiate:

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

Evaluate at $x = 16$ :

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

$= \frac{1}{8}$

Use the linearization formula:

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

3. Approximate the desired function value

Plug in $x = 15$ :

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

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

$= 3.875$

A calculator gives $15 \approx 3.872 \dots$ , so the estimate is close.

Curvature

Linear approximations get better as $x$ gets closer to $a$ . Accuracy also depends on how much the function curves near $x = a$ .

If $f^{''} (x) > 0$ , then $f (x)$ is concave up (shaped “like a cup”).
- The tangent line lies below $f (x)$ , so the linear approximation will underestimate the actual value.
If $f^{''} (x) < 0$ , then $f (x)$ is concave down (shaped “like a frown”).
- The tangent line lies above $f (x)$ , so the linear approximation will overestimate the actual value.

The image below shows this relationship between tangent lines and concavity.

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

Solution

(spoiler)

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

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

A convenient known point is $x = 1$ because

$ln (1) = 0$

So the known point is $(1, 0)$ .

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

Differentiate:

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

Evaluate at $x = 1$ :

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

Write the tangent line at $(1, 0)$ :

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

So the linearization at $x = 1$ is

$L (x) = x - 1$

3. Approximate the desired function value

Plug in $x = 1.1$ :

$L (1.1) = 1.1 - 1$

$= 0.1$

To decide whether this is an overestimate or underestimate, look at concavity:

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

Since $f^{''} (x) < 0$ for all $x > 0$ , $ln (x)$ is concave down everywhere in its domain. That means the tangent line lies above the curve, so the tangent line approximation is an overestimate.

A calculator confirms that $ln (1.1) \approx 0.095 \dots$ , which is slightly less than $0.1$ .

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

Solution

(spoiler)

Let $f (x) = e^{x}$ .

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

Since $0.02$ is close to $0$ , choose $a = 0$ . Then

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

so the known point is $(0, 1)$ .

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

Differentiate:

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

Evaluate at $x = 0$ :

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

Write the linearization at $x = 0$ :

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

$= 1 + x$

3. Approximate the desired function value

Plug in $x = 0.02$ :

$L (0.02) = 1 + 0.02$

$= 1.02$

To decide whether this is an overestimate or underestimate, check concavity:

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

Because $e^{x} > 0$ for all $x$ , $f (x) = e^{x}$ is always concave up. That means the tangent line lies below the curve, so the linear approximation is an underestimate.

A calculator confirms that $e^{0.02} \approx 1.0202 \dots$ , which is slightly greater than $1.02$ .

Local linear approximation (tangent line approximation)

Uses tangent line at $x = a$ to estimate $f (x)$ near $a$
Linearization formula: $L (x) = f (a) + f^{'} (a) (x - a)$
Good for quick estimates when $x$ is close to $a$

Example: Estimating $2.0 1^{3}$

Choose $f (x) = x^{3}$ , $a = 2$
$f (2) = 8$ , $f^{'} (2) = 12$
Linearization: $L (x) = 8 + 12 (x - 2)$
$L (2.01) = 8.12$ (close to actual $8.120601$ )

Example: Estimating $15$

Choose $f (x) = x$ , $a = 16$
$f (16) = 4$ , $f^{'} (16) = \frac{1}{8}$
Linearization: $L (x) = 4 + \frac{1}{8} (x - 16)$
$L (15) = 3.875$ (close to actual $3.872$ )

Curvature and concavity

Accuracy improves as $x$ approaches $a$
Concavity determined by $f^{''} (x)$ :
- $f^{''} (x) > 0$ : concave up, tangent line underestimates
- $f^{''} (x) < 0$ : concave down, tangent line overestimates

Example: Estimating $ln (1.1)$

$f (x) = ln (x)$ , $a = 1$
$f (1) = 0$ , $f^{'} (1) = 1$
Linearization: $L (x) = x - 1$
$L (1.1) = 0.1$ (overestimate, since $f^{''} (x) < 0$ )

Example: Estimating $e^{0.02}$

$f (x) = e^{x}$ , $a = 0$
$f (0) = 1$ , $f^{'} (0) = 1$
Linearization: $L (x) = 1 + x$
$L (0.02) = 1.02$ (underestimate, since $f^{''} (x) > 0$ )

Linear approximations

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 $L (x)$

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

This is the tangent line to $f (x)$ at $x = a$ . If $x_{1}$ is close to $a$ , then $L (x_{1})$ is an estimate of the actual value $f (x_{1})$ .

Example

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

Solution

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

Define

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

Since $2.01$ is close to $2$ , choose $a = 2$ . Then

$f (2) = 8$

so the known point is $(2, 8)$ .

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

Differentiate:

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

Evaluate the slope at $x = 2$ :

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

Use point-slope form at $(2, 8)$ :

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

Solve for $y$ :

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

Replace $y$ with $L (x)$ to write the linearization at $x = 2$ :

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

3. Approximate the desired function value

Plug in $x = 2.01$ :

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

$= 8 + 12 (0.01)$

$= 8.12$

The actual value (using a calculator) is $8.120601$ , so the linear approximation is very close.

Estimate $15$ with a linear approximation.

Solution

(spoiler)

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

Let

$f (x) = x$

Since $15$ is close to $16$ and $16$ is easy to compute, choose $a = 16$ . Then

$f (16) = 4$

so the known point is $(16, 4)$ . This also tells us $15$ should be a little less than $4$ .

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

Differentiate:

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

Evaluate at $x = 16$ :

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

$= \frac{1}{8}$

Use the linearization formula:

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

3. Approximate the desired function value

Plug in $x = 15$ :

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

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

$= 3.875$

A calculator gives $15 \approx 3.872 \dots$ , so the estimate is close.

Curvature

Linear approximations get better as $x$ gets closer to $a$ . Accuracy also depends on how much the function curves near $x = a$ .

If $f^{''} (x) > 0$ , then $f (x)$ is concave up (shaped “like a cup”).
- The tangent line lies below $f (x)$ , so the linear approximation will underestimate the actual value.
If $f^{''} (x) < 0$ , then $f (x)$ is concave down (shaped “like a frown”).
- The tangent line lies above $f (x)$ , so the linear approximation will overestimate the actual value.

The image below shows this relationship between tangent lines and concavity.

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

Solution

(spoiler)

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

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

A convenient known point is $x = 1$ because

$ln (1) = 0$

So the known point is $(1, 0)$ .

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

Differentiate:

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

Evaluate at $x = 1$ :

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

Write the tangent line at $(1, 0)$ :

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

So the linearization at $x = 1$ is

$L (x) = x - 1$

3. Approximate the desired function value

Plug in $x = 1.1$ :

$L (1.1) = 1.1 - 1$

$= 0.1$

To decide whether this is an overestimate or underestimate, look at concavity:

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

Since $f^{''} (x) < 0$ for all $x > 0$ , $ln (x)$ is concave down everywhere in its domain. That means the tangent line lies above the curve, so the tangent line approximation is an overestimate.

A calculator confirms that $ln (1.1) \approx 0.095 \dots$ , which is slightly less than $0.1$ .

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

Solution

(spoiler)

Let $f (x) = e^{x}$ .

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

Since $0.02$ is close to $0$ , choose $a = 0$ . Then

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

so the known point is $(0, 1)$ .

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

Differentiate:

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

Evaluate at $x = 0$ :

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

Write the linearization at $x = 0$ :

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

$= 1 + x$

3. Approximate the desired function value

Plug in $x = 0.02$ :

$L (0.02) = 1 + 0.02$

$= 1.02$

To decide whether this is an overestimate or underestimate, check concavity:

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

Because $e^{x} > 0$ for all $x$ , $f (x) = e^{x}$ is always concave up. That means the tangent line lies below the curve, so the linear approximation is an underestimate.

A calculator confirms that $e^{0.02} \approx 1.0202 \dots$ , which is slightly greater than $1.02$ .