4.4: Linear Approximations
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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