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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 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).

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

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