4.4 Linear approximations

Achievable AP Calculus AB

4. Contextual uses

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Linear approximations

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What you’ll learn

Use the tangent line equation to approximate nearby function values
Use the 2nd derivative $(f^{''})$ to justify whether an approximation over- or underestimates the actual value

When a function is differentiable, its curve looks almost perfectly straight if you zoom in close enough to a single point. Because of this, the equation of a tangent line at a known point can be used approximate the values of the function for nearby inputs. This is called local linear approximation or tangent line approximation.

The Linearization formula:

The linear approximation function, $L (x)$ , to a curve $f (x)$ at $x = a$ is given by:

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

This formula is simply the equation of the tangent line to $f (x)$ at $x = a$ . If an input $x_{1}$ is close to the center of approximation $a$ , then $L (x_{1})$ is a highly accurate estimate of the actual value $f (x_{1})$ .

Example

What is the approximate value of $f (x) = e^{1 - x}$ at $x = 0.8$ using the tangent line to the graph at $x = 1$ ?

Bonus problem: Using the tangent line to the graph at $x = 0$ instead, what is the approximate value of the same function at $x = 0.8$ ?

Solution

(spoiler)

This question essentially asks for the estimate of $f (0.8) = e^{0.2}$ .

1. Find the point of tangency and the slope.

Point: $f (1) = e^{0} = 1$
Slope: $f^{'} (x) = - e^{1 - x} ⟹ f^{'} (1) = - 1$

2. Write the tangent line equation.

$y y = f (1) + f^{'} (1) (x - 1) = 1 - 1 (x - 1)$

3. Compute the approximation $L (x)$ .

Substitute $x = 0.8$ :

$L (0.8) \approx 1 - 1 (0.8 - 1) = 1.2$

In fact, the actual value is $e^{0.2} \approx 1.221$ , so the linear estimate is very close.

Bonus problem solution

(spoiler)

To write the tangent line equation to $f (x) = e^{1 - x}$ at $x = 0$ instead, find the point and the slope:

Point: $f (0) = e^{1}$
Slope: $f^{'} (x) = - e^{1 - x} ⟹ f^{'} (0) = - e$

Then the equation of the tangent line is

$y y = f (0) + f^{'} (0) (x - 0) = e - e x$

The approximation for $f (0.8)$ using this line instead is:

$L (0.8) \approx e - e (0.8) = 0.544$

Notice how much further this is from the actual value of $e^{0.2} \approx 1.221$ . Because the target value we chose to estimate ( $x = 0.8$ ) is further away from the anchor point where the tangent line was built ( $a = 0$ ), the approximation is significantly less accurate.

Error analysis: Over vs. understimates

As demonstrated in the bonus problem, the accuracy of a linear approximation improves as the input $x$ gets closer to the anchor point $a$ .

Mathematically, error is defined as the difference between the actual value and the approximation:

$Error = ∣ f (x) - L (x) ∣$

By analyzing how $f (x)$ is shaped near $x = a$ , we can determine whether the approximation $L (x)$ is less than or greater than the actual value $f (x)$ .

The 2nd 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 underestimates 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 overestimates the actual value.

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

Figure 4.4.2 Tangent lines and concavity

Example

Let $f$ be the function defined by $f (x) = ln (x)$ .

a) Use the line tangent to the graph of $f$ at $x = 1$ to approximate $f (1.1)$ .

b) Determine whether this approximation is an overestimate or an underestimate of the actual value of $f (1.1)$ . Justify your answer.

Solutions

a) Use the line tangent to the graph of $f$ at $x = 1$ to approximate $f (1.1)$ .

(spoiler)

1. Write the tangent line at $x = 1$ .

Point: $f (1) = ln (1) = 0$
Slope: $f^{'} (x) = \frac{1}{x} ⟹ f^{'} (1) = 1$

So the tangent line equation at $x = 1$ is

$y = x - 1$

2. Approximate $f (1.1)$ .

Plug in $x = 1.1$ :

$L (1.1) \approx 1.1 - 1 = 0.1$

b) Overestimate or underestimate?

(spoiler)

To decide whether this is an overestimate or underestimate, find the 2nd derivative:

$f^{''} (x) = - \frac{1}{x ^{2}}, x > 0$

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

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

Challenge problem

Let $f$ be a twice-differentiable function. Selected values of $f$ and its derivative $f^{'}$ are given in the table below.

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

$1$ $3$ $- 4$

$2$ $- 1$ $- 2$

a) Must there be a value $c$ , for $1 < c < 2$ , such that $f (c) = 0$ ? Justify your answer.

b) Write an equation for the line tangent to the graph of $f$ at $x = 1$ . Use this line to approximate the zero of $f$ .

$x$	$f (x)$	$f^{'} (x)$
$1$	$3$	$- 4$
$2$	$- 1$	$- 2$

Solutions

a) Must there be a value $c$ , for $1 < c < 2$ , such that $f (c) = 0$ ?

(spoiler)

The wording of this problem suggests a textbook application of the Intermediate value theorem, where the target value $L = 0$ . Your justification must state that the conditions are met:

Continuity:
- Since $f$ is twice-differentiable, and differentiability implies continuity, $f$ is guaranteed to be continuous on $[1, 2]$ .
$L$ is between $f (a)$ and $f (b)$ :
- From the table, $f (1) = 3$ and $f (2) = - 1$
- Since $- 1 < 0 < 3$ , this condition has been met.

Therefore, the Intermediate value theorem guarantees a $c$ in the open interval $(1, 2)$ such that $f (c) = 0$ .

b) Write an equation for the line tangent to the graph of $f$ at $x = 1$ . Use this line to approximate the zero of $f$ .

(spoiler)

The tangent line equation at $x = 1$ is

$y y = f (1) + f^{'} (1) (x - 1) = 3 - 4 (x - 1)$

A zero, or root, of $f$ is where the function crosses the $x$ -axis, meaning $f (x) = 0$ . While a tangent line is usually used to approximate a $y$ -value, here we work “backwards” to find the value of $x$ for which $y = 0$ :

$0 4 (x - 1) 4 x - 4 x = 3 - 4 (x - 1) = 3 = 3 = \frac{7}{4}$

Therefore, the tangent line approximates the zero of $f$ to be at $x = 1.75$ . Notice how this approximation falls perfectly within the open interval $(1, 2)$ guaranteed by the IVT in part (a).

Local Linear Approximation

Tangent line at a known point used to estimate nearby function values
Linearization formula: $L (x) = f (a) + f^{'} (a) (x - a)$
Accuracy decreases as input $x$ moves farther from anchor point $a$

Error Analysis: Over vs. Underestimates

Error = $∣ f (x) - L (x) ∣$
If $f^{''} (x) > 0$ (concave up): tangent line lies below curve → approximation underestimates
If $f^{''} (x) < 0$ (concave down): tangent line lies above curve → approximation overestimates

Applying the Linearization Formula

Find point $(a, f (a))$ and slope $f^{'} (a)$ , then build tangent line equation
Substitute target $x$ value into $L (x)$ for the approximation
Can solve “backwards” (set $L (x) = 0$ ) to approximate zeros of $f$

Linear approximations

What you’ll learn

Use the tangent line equation to approximate nearby function values
Use the 2nd derivative $(f^{''})$ to justify whether an approximation over- or underestimates the actual value

The Linearization formula:

The linear approximation function, $L (x)$ , to a curve $f (x)$ at $x = a$ is given by:

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

Example

What is the approximate value of $f (x) = e^{1 - x}$ at $x = 0.8$ using the tangent line to the graph at $x = 1$ ?

Bonus problem: Using the tangent line to the graph at $x = 0$ instead, what is the approximate value of the same function at $x = 0.8$ ?

Solution

(spoiler)

This question essentially asks for the estimate of $f (0.8) = e^{0.2}$ .

1. Find the point of tangency and the slope.

Point: $f (1) = e^{0} = 1$
Slope: $f^{'} (x) = - e^{1 - x} ⟹ f^{'} (1) = - 1$

2. Write the tangent line equation.

$y y = f (1) + f^{'} (1) (x - 1) = 1 - 1 (x - 1)$

3. Compute the approximation $L (x)$ .

Substitute $x = 0.8$ :

$L (0.8) \approx 1 - 1 (0.8 - 1) = 1.2$

In fact, the actual value is $e^{0.2} \approx 1.221$ , so the linear estimate is very close.

Bonus problem solution

(spoiler)

To write the tangent line equation to $f (x) = e^{1 - x}$ at $x = 0$ instead, find the point and the slope:

Point: $f (0) = e^{1}$
Slope: $f^{'} (x) = - e^{1 - x} ⟹ f^{'} (0) = - e$

Then the equation of the tangent line is

$y y = f (0) + f^{'} (0) (x - 0) = e - e x$

The approximation for $f (0.8)$ using this line instead is:

$L (0.8) \approx e - e (0.8) = 0.544$

Error analysis: Over vs. understimates

As demonstrated in the bonus problem, the accuracy of a linear approximation improves as the input $x$ gets closer to the anchor point $a$ .

Mathematically, error is defined as the difference between the actual value and the approximation:

$Error = ∣ f (x) - L (x) ∣$

By analyzing how $f (x)$ is shaped near $x = a$ , we can determine whether the approximation $L (x)$ is less than or greater than the actual value $f (x)$ .

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 underestimates 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 overestimates the actual value.

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

Example

Let $f$ be the function defined by $f (x) = ln (x)$ .

a) Use the line tangent to the graph of $f$ at $x = 1$ to approximate $f (1.1)$ .

b) Determine whether this approximation is an overestimate or an underestimate of the actual value of $f (1.1)$ . Justify your answer.

Solutions

a) Use the line tangent to the graph of $f$ at $x = 1$ to approximate $f (1.1)$ .

(spoiler)

1. Write the tangent line at $x = 1$ .

Point: $f (1) = ln (1) = 0$
Slope: $f^{'} (x) = \frac{1}{x} ⟹ f^{'} (1) = 1$

So the tangent line equation at $x = 1$ is

$y = x - 1$

2. Approximate $f (1.1)$ .

Plug in $x = 1.1$ :

$L (1.1) \approx 1.1 - 1 = 0.1$

b) Overestimate or underestimate?

(spoiler)

To decide whether this is an overestimate or underestimate, find the 2nd derivative:

$f^{''} (x) = - \frac{1}{x ^{2}}, x > 0$

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

Challenge problem

Let $f$ be a twice-differentiable function. Selected values of $f$ and its derivative $f^{'}$ are given in the table below.

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

$1$ $3$ $- 4$

$2$ $- 1$ $- 2$

a) Must there be a value $c$ , for $1 < c < 2$ , such that $f (c) = 0$ ? Justify your answer.

b) Write an equation for the line tangent to the graph of $f$ at $x = 1$ . Use this line to approximate the zero of $f$ .

$x$	$f (x)$	$f^{'} (x)$
$1$	$3$	$- 4$
$2$	$- 1$	$- 2$

Solutions

a) Must there be a value $c$ , for $1 < c < 2$ , such that $f (c) = 0$ ?

(spoiler)

The wording of this problem suggests a textbook application of the Intermediate value theorem, where the target value $L = 0$ . Your justification must state that the conditions are met:

Continuity:
- Since $f$ is twice-differentiable, and differentiability implies continuity, $f$ is guaranteed to be continuous on $[1, 2]$ .
$L$ is between $f (a)$ and $f (b)$ :
- From the table, $f (1) = 3$ and $f (2) = - 1$
- Since $- 1 < 0 < 3$ , this condition has been met.

Therefore, the Intermediate value theorem guarantees a $c$ in the open interval $(1, 2)$ such that $f (c) = 0$ .

b) Write an equation for the line tangent to the graph of $f$ at $x = 1$ . Use this line to approximate the zero of $f$ .

(spoiler)

The tangent line equation at $x = 1$ is

$y y = f (1) + f^{'} (1) (x - 1) = 3 - 4 (x - 1)$

$0 4 (x - 1) 4 x - 4 x = 3 - 4 (x - 1) = 3 = 3 = \frac{7}{4}$

Therefore, the tangent line approximates the zero of $f$ to be at $x = 1.75$ . Notice how this approximation falls perfectly within the open interval $(1, 2)$ guaranteed by the IVT in part (a).

Achievable AP Calculus AB

4. Contextual uses

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.