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4.4: Linear Approximations

Difficulty Level: At Grade Created by: CK-12
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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 \begin{align*}a\end{align*} be the point where the tangent touches the graph, \begin{align*}L(x)\end{align*} be the tangent, and \begin{align*}f(x)\end{align*} be the function.

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

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

Draw horizontal lines from \begin{align*}a\end{align*}, \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 \begin{align*}y-\end{align*}axis: \begin{align*}f(a)\end{align*}, \begin{align*}f(x)\end{align*}, \begin{align*}L(x)\end{align*}. Label them.

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

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

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

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

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

\begin{align*}p\end{align*} distance \begin{align*}pq\end{align*} linear approx.of \begin{align*}f1(q)\end{align*} real valueof \begin{align*}f1(q)\end{align*} error underestimation/overestimation
\begin{align*}x = -0.2\end{align*}
\begin{align*}x = -0.5\end{align*}
\begin{align*}x = -0.6\end{align*}
\begin{align*}x = -1.2\end{align*}
  • At what \begin{align*}x-\end{align*}value(s) is the error less than \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 \begin{align*}f1(x) = x^3 - 3x^2 - 2x + 6\end{align*}. Evaluate it at \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 \begin{align*}L(x)=\end{align*}
  • What is \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 \begin{align*}f1(x) = x^3 - 3x^2 - 2x + 6\end{align*} and place a tangent line \begin{align*}a = 1\end{align*}.

  • If you were to draw a point \begin{align*}p\end{align*} on the graph to the left of \begin{align*}a = 1\end{align*}, is the approximation an overestimate or an underestimate?
  • If you draw a point \begin{align*}p\end{align*} on the graph to the right of \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 \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*}?

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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TI.MAT.ENG.SE.1.Calculus.4.4