Approximating Values of a Function Using Local Linearity and Linearization

Hannah Hill

7 min read

Next Topic - Using L'Hopitals Rule for Determining Limits in Indeterminate Forms

Listen to this study note

Study Guide Overview

This study guide covers local linearity and linearization to approximate function values using tangent lines. It explains how to use the point-slope form and linearization formula to construct tangent line equations. The guide also discusses how to determine overestimations and underestimations based on concavity. Finally, it provides practice questions and solutions, including AP-style examples, covering how to find tangent lines, approximate function values, and determine over/underestimation.

#Approximating Function Values with Local Linearity and Linearization 🚀

Imagine zooming in super close on a curve until it looks almost perfectly straight. That's the idea behind local linearity! We use tangent lines to approximate function values near a known point. It's like using a straight path to guess where a winding trail goes. Let's dive in!

#↗️ Linearization and Tangent Line Approximation

Remember tangent lines from Unit 2? The slope of the line tangent to a graph at a point is the function's derivative at that point. We use this slope and the point's coordinates to build the tangent line equation. This equation is our linearization.

#Point-Slope Form

The most reliable way to build the tangent line equation is using the point-slope formula:

$y - y_1 = m(x - x_1)$

Where:

$(x_1, y_1)$ is the point of tangency
$m$ is the slope of the tangent line (the derivative at $x_1$ )

#Linearization Formula

While you might see the linearization formula, it's just a fancy version of point-slope. You can use it if you like, but point-slope is your friend!

$L(x) = f(a) + f'(a)(x - a)$

$L(x)$ is the linearization of $f(x)$ at $x=a$
$f(a)$ is the function value at $x=a$
$f'(a)$ is the derivative of the function at $x=a$

Key Concept

Essentially, both formulas do the same thing: they build a line that closely approximates the function near a specific point. Point-slope is often easier to remember!

#How to Approximate

Find the point of tangency: $(x_1, y_1)$
Calculate the derivative $f'(x)$ and evaluate it at $x_1$ to find the slope, $m = f'(x_1)$ .
Build the tangent line equation using the point-slope form or the linearization formula.
Plug in the x-value you want to approximate into the tangent line equation and solve for y. This y-value is your approximation.

Quick Fact

Linearization is most accurate when approximating values close to the point of tangency. The further you go, the less accurate it becomes.

#🔀 Under/Overestimations

AP loves to ask if your approximation is an overestimate or underestimate! Here's the secret:

Concave Up: If the function is concave up at the point of tangency, the tangent line lies below the curve, making it an underestimate. Think of a smile 😊 - the line is below the smile.

markdown-image

Graph of $x^2$ and corresponding tangent line ( $y-1=2(x-1)$ )

Concave Down: If the function is concave down at the point of tangency, the tangent line lies above the curve, making it an overestimate. Think of a frown 🙁 - the line is above the frown.

markdown-image

Graph of $-x^2$ and corresponding tangent line ( $y+1=2(x+1)$ )

Memory Aid

Concave Up = Underestimate (think: the tangent line is under the curve, like a cup) Concave Down = Overestimate (think: the tangent line is over the curve, like a frown)

#Linearization Practice Questions

Let's tackle a real AP-style question!

Practice Question

#2010 AP Calculus AB #6a

a) Write an equation for the line tangent to the graph of $y=f(x)$ at x = 1.

Solution:

Find the slope: Given $\frac{dy}{dx} = xy^3$ and $f(1) = 2$ , plug in $x=1$ and $y=2$ to find the slope at $x=1$ . $f'(1) = (1)(2)^3 = 8$
Use point-slope form: The tangent line equation is: $y - 2 = 8(x - 1)$

#2010 AP Calculus AB #6b

b) Use the tangent line equation from part (a) to approximate $f(1.1)$ . Given that $f(x)>0$ for $1<x<1.1$ , is the approximation for $f(1.1)$ greater than or less than $f(1.1)$ ? Explain your reasoning.

Solution:

Approximate: Plug $x = 1.1$ into the tangent line equation: $y - 2 = 8(1.1 - 1)$ $y - 2 = 8(0.1)$ $y = 2 + 0.8$ $f(1.1) \approx 2.8$
Over/Underestimate: Since $f(x) > 0$ for $1 < x < 1.1$ , we know the second derivative is positive, indicating that the function is concave up. Therefore, our approximation is an underestimate.

markdown-image

Graph of the function and its tangent line

#Multiple Choice Practice

The function $f$ is differentiable, and the line $y=3x-2$ is tangent to the graph of $f$ at $x=1$ . What is the value of $f(1) + f'(1)$ ? (A) 1 (B) 2 (C) 3 (D) 4 (E) 5
Let $f$ be a differentiable function with $f(2) = 5$ and $f'(2) = -3$ . Using the tangent line approximation, what is the approximate value of $f(2.1)$ ? (A) 4.7 (B) 4.8 (C) 4.9 (D) 5.1 (E) 5.3

#Free Response Question

Let $f$ be a function that is differentiable on the interval $(0, 10)$ . The function $f$ has values given in the table below.

x	2	4	6	8	10
f(x)	1	3	4	2	0

(a) Use the data in the table to approximate $f'(3)$ . Show the computations that lead to your answer.

(b) Use a tangent line approximation to estimate $f(4.5)$ .

Answer Key:

MCQ 1: (D) 4, since $f(1)=3(1)-2=1$ and $f'(1)=3$ , so $f(1)+f'(1)=1+3=4$
MCQ 2: (A) 4.7, since $L(x) = 5 - 3(x-2)$ , so $f(2.1) \approx 5 - 3(2.1-2) = 5 - 3(0.1) = 4.7$
FRQ: (a) $f'(3) \approx \frac{f(4)-f(2)}{4-2} = \frac{3-1}{2} = 1$ (b) The tangent line at x=4 is $y-3 = f'(4)(x-4)$ . Using the slope between (2,1) and (6,4), we can estimate $f'(4) \approx \frac{4-1}{6-2} = \frac{3}{4}$ . Then, $y-3 = \frac{3}{4}(x-4)$ . $f(4.5) \approx 3 + \frac{3}{4}(4.5-4) = 3 + \frac{3}{4}(0.5) = 3 + \frac{3}{8} = 3.375$ (c) We don't have enough information to determine if the approximation is an over or underestimate. We would need information about the concavity of $f$ near $x=4$ .

#🔷 Closing

Awesome job! 🎉

You've now mastered the art of using tangent lines to approximate function values and can determine if those approximations are over or underestimates. This is a key skill that pops up in many AP problems, so make sure you've got it down! Keep up the great work! 🙌

Linear approximation is a foundational concept that often appears in conjunction with other topics like related rates and optimization. It's crucial to understand this topic well!

Exam Tip

When asked to justify if an approximation is an over or underestimate, always refer to the concavity of the function at the point of tangency. This is a must-have for full credit!

Common Mistake

A common mistake is confusing the function value with the derivative value. Remember, the derivative is the slope of the tangent line, not the y-value of the function.