1 / 5

Linear Approximation

Linear Approximation. Katherine Han Period 3 4-29-03. Concept of Linear Approximation. Given a certain equation, f(x), try to find the equation of the tangent line. After finding the equation of the tangent: Substitute “x” in order to get a linear approximation of “y.”.

cmoffitt
Download Presentation

Linear Approximation

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Linear Approximation Katherine Han Period 3 4-29-03

  2. Concept of Linear Approximation • Given a certain equation, f(x), try to find the equation of the tangent line. • After finding the equation of the tangent: • Substitute “x” in order to get a linear approximation of “y.”

  3. Using Linear Approximation Given f(x) • Step 1: Find a good “x” value that would be easy to work with, that is near the actual value given – this value will be called “a” • Step 2: dx = real value – a • Step 3: dy = f’(a)*dx • Step 4: f(x) = f(a) + dy • Step 5: Compare to the actual value to check to see that it is close

  4. Example Problem Use Linearization to estimate f(1.99) of f(x)=2x3 -4x2+1 • Step 1: Choose a nice value close to 1.99 to be your “a” • a = 2 • Step 2: dx = real value – a • dx = 1.99-2 = -0.01 • Step 3: dy=f’(a)*dx = (24-16)*(-.01) = -.08 • Step 4: f(1.99)=f(a)+dy=f(2)+dy=1- 0.08=.92 • Step 5: Compare to actual value f(1.99)=2x3 –4x2+1 =.920798

  5. Yet Another Example If f(3) =8, f’(3) = -4, then f(3.02) • a= 3 • dx = 3.02 – 3 = .02 • dy = f’(a)*dx • -4*.02 = -.08 • f(x) = f(a) + dy • = 8+(-.08) = 7.92

More Related