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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.”.
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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.”
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
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
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