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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\begin{align*}a\end{align*} be the point where the tangent touches the graph, L(x)\begin{align*}L(x)\end{align*} be the tangent, and f(x)\begin{align*}f(x)\end{align*} be the function.

On the picture, the point x\begin{align*}x\end{align*} is the x\begin{align*}x-\end{align*} coordinate of the vertical line.

Draw a vertical line from a\begin{align*}a\end{align*} to the x\begin{align*}x-\end{align*}axis.

Draw horizontal lines from a\begin{align*}a\end{align*}, f(x)\begin{align*}f(x)\end{align*}, and the intersection of the vertical line with the tangent line.

At this stage, you should have three points on the y\begin{align*}y-\end{align*}axis: f(a)\begin{align*}f(a)\end{align*}, f(x)\begin{align*}f(x)\end{align*}, L(x)\begin{align*}L(x)\end{align*}. Label them.

• Which of these points can you use to represent the estimate, or linear approximation, of f(x)\begin{align*}f(x)\end{align*} near a\begin{align*}a\end{align*}?
• 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)=x33x22x+6\begin{align*}f1(x) = x^3 - 3x^2 - 2x + 6\end{align*} is shown. The tangent line at a=1\begin{align*}a = -1\end{align*} is L(x)=7x+11\begin{align*}L(x) = 7x + 11\end{align*}. The trace object q\begin{align*}q\end{align*} is at the point (0.2, 5.488).

If you draw a vertical line from the x\begin{align*}x-\end{align*}axis through this point, you will get the point p=(0.212658,12.4886)\begin{align*}p = (0.212658, 12.4886)\end{align*}.

The distance pq=7.039997\begin{align*}pq = 7.039997\end{align*} or 7.04.

• What numerical value represents the linear approximation of f1(q)\begin{align*}f1(q)\end{align*} near a=1\begin{align*}a = -1\end{align*}?
• What numerical value gives the error associated with this linear approximation?
• What is the true value of f1(q)\begin{align*}f1(q)\end{align*}?
• Is this an underestimation or an overestimation?

Repeat the process above and complete the table below for the x\begin{align*}x-\end{align*}values given.

p\begin{align*}p\end{align*} distance pq\begin{align*}pq\end{align*} linear approx.of f1(q)\begin{align*}f1(q)\end{align*} real valueof f1(q)\begin{align*}f1(q)\end{align*} error underestimation/overestimation
x=0.2\begin{align*}x = -0.2\end{align*}
x=0.5\begin{align*}x = -0.5\end{align*}
x=0.6\begin{align*}x = -0.6\end{align*}
x=1.2\begin{align*}x = -1.2\end{align*}
• At what x\begin{align*}x-\end{align*}value(s) is the error less than 0.5\begin{align*}0.5\end{align*}?
• 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)=x33x22x+6\begin{align*}f1(x) = x^3 - 3x^2 - 2x + 6\end{align*}. Evaluate it at x=1\begin{align*}x = -1\end{align*}. 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)=\begin{align*}L(x)=\end{align*}
• What is L(1.03)\begin{align*}L(-1.03)\end{align*}? What does this value represent?
• Calculate the error with this estimate.

## Part 3 – Underestimates versus overestimates

Graph the function f1(x)=x33x22x+6\begin{align*}f1(x) = x^3 - 3x^2 - 2x + 6\end{align*} and place a tangent line a=1\begin{align*}a = 1\end{align*}.

• If you were to draw a point p\begin{align*}p\end{align*} on the graph to the left of a=1\begin{align*}a = 1\end{align*}, is the approximation an overestimate or an underestimate?
• If you draw a point p\begin{align*}p\end{align*} on the graph to the right of a=1\begin{align*}a=1\end{align*}, 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\begin{align*}x\end{align*} be for the linear approximation of f1(x)=x33x22x+6\begin{align*}f1(x) = x^3 - 3x^2 - 2x + 6\end{align*} at a=1\begin{align*}a = -1\end{align*} to be within 0.2 units of the true value of f1(x)\begin{align*}f1(x)\end{align*}?

Graph \begin{align*}f2 = f1 + 0.2\end{align*} and \begin{align*}f3 = f1 - 0.2\end{align*} with \begin{align*}f1\end{align*} and the tangent line.

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

Now we want to ask the same questions when the point of tangency is at \begin{align*}a = 1\end{align*}.

• 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 \begin{align*}a = 1 \end{align*} is accurate to within 0.2 units of \begin{align*}f1(x)\end{align*}.

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