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# 4.4: Linear Approximations

Difficulty Level: At Grade Created by: CK-12

This activity is intended to supplement Calculus, Chapter 3, Lesson 8.

## Part 1 – Introduction

Linear approximation uses a tangent line to estimate values of a function near the point of tangency. For this reason, linear approximation is also referred to as tangent line approximation.

On the graph to the right, let $a$ be the point where the tangent touches the graph, $L(x)$ be the tangent, and $f(x)$ be the function.

On the picture, the point $x$ is the $x-$ coordinate of the vertical line.

Draw a vertical line from $a$ to the $x-$axis.

Draw horizontal lines from $a$, $f(x)$, and the intersection of the vertical line with the tangent line.

At this stage, you should have three points on the $y-$axis: $f(a)$, $f(x)$, $L(x)$. Label them.

• Which of these points can you use to represent the estimate, or linear approximation, of $f(x)$ near $a$?
• How can you use these labels to represent the error associated with this estimate?
• Is this estimate an overestimate or an underestimate? Explain.

## Part 2 – Investigating linear approximation

On the graph at the right, $f1(x) = x^3 - 3x^2 - 2x + 6$ is shown. The tangent line at $a = -1$ is $L(x) = 7x + 11$. The trace object $q$ is at the point (0.2, 5.488).

If you draw a vertical line from the $x-$axis through this point, you will get the point $p = (0.212658, 12.4886)$.

The distance $pq = 7.039997$ or 7.04.

• What numerical value represents the linear approximation of $f1(q)$ near $a = -1$?
• What numerical value gives the error associated with this linear approximation?
• What is the true value of $f1(q)$?
• Is this an underestimation or an overestimation?

Repeat the process above and complete the table below for the $x-$values given.

$p$ distance $pq$ linear approx.of $f1(q)$ real valueof $f1(q)$ error underestimation/overestimation
$x = -0.2$
$x = -0.5$
$x = -0.6$
$x = -1.2$
• At what $x-$value(s) is the error less than $0.5$?
• What do you notice about graph of the function and the graph of the tangent line as you get close to the point of tangency?
• Based on your observations, explain why the relationship between a tangent and a graph at the point of tangency is often referred to as local linearization.

Typically, you will have a function but not a graph to find the linear approximation.

• Find the derivative of $f1(x) = x^3 - 3x^2 - 2x + 6$. Evaluate it at $x = -1$. This is the slope of the line. Use the slope and the point (-1, 4) to get the equation of the line.
• The tangent line $L(x)=$
• What is $L(-1.03)$? What does this value represent?
• Calculate the error with this estimate.

## Part 3 – Underestimates versus overestimates

Graph the function $f1(x) = x^3 - 3x^2 - 2x + 6$ and place a tangent line $a = 1$.

• If you were to draw a point $p$ on the graph to the left of $a = 1$, is the approximation an overestimate or an underestimate?
• If you draw a point $p$ on the graph to the right of $a=1$, is the approximation an overestimate or an underestimate?
• What is the significance of the point of tangency?
• Generalize your findings about when a linear approximation produces an overestimate and when it produces an underestimate.

## Part 4 –Finding intervals of accuracy

How close to -1 must $x$ be for the linear approximation of $f1(x) = x^3 - 3x^2 - 2x + 6$ at $a = -1$ to be within 0.2 units of the true value of $f1(x)$?

Graph $f2 = f1 + 0.2$ and $f3 = f1 - 0.2$ with $f1$ and the tangent line.

• How would you use the graphs to answer the question posed in this problem?
• How close to -1 must $x$ be for the linear approximation of $f1(x) = x^3 - 3x^2 - 2x + 6$ at $a = -1$ to be within 0.2 units of the true value of $f1(x)$?

Now we want to ask the same questions when the point of tangency is at $a = 1$.

• How does this situation differ from the one we just had?
• Use graphical or algebraic methods to find an interval that ensures the linear approximation at $a = 1$ is accurate to within 0.2 units of $f1(x)$.

Feb 23, 2012

Nov 04, 2014