Linear Approximation to ln x at x=1 | Massachusetts Institute of Technology - Edubirdie

University:
Massachusetts Institute of Technology
Course:
18.01SC | Single Variable Calculus
Academic year:

2024
Views:

480

Pages:

1
Author:

Will Abbott

Linear Approximation to ln x at x = 1 If you have a curve y = f (x), it is approximately the same as its tangent line y = f (x0 ) + f � (x0 )(x − x0 ). y y = b+a(x-x0) b = f(x0) ; a = f’(x0 ) y=f(x) (x0 ,f(x0 )) x Figure 1: Tangent as a linear approximation to a curve The tangent line approximates f (x). It gives a good approximation near the tangent point x0 . As you move away from x0 , however, the approximation grows less accurate. f (x) ≈ f (x0 ) + f � (x0 )(x − x0 ) Example 1 Let f (x) = ln x. Then f � (x) = x1 . We’ll use the base point x0 = 1 because we can easily evaluate ln 1 = 0. Note also that f � (x0 ) = 11 = 1. Then the formula for linear approximation tells us that: f (x) ≈ f (x0 ) + f � (x0 )(x − x0 ) ln x ≈ ln(1) + 1(x − 1) ln x ≈ 0 + (x − 1) ln x ≈ (x − 1) Graph the curve y = ln x and the line y = x − 1. You’ll see that the two graphs are very close together when x = x0 = 1. You’ll also see that they’re only near each other when x is near 1. The point of linear approximation is that the curve (in this case y = ln x) is approximately the same as the tangent line (y = x − 1) when x is close to the base point x0 .

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