Find the linear approximation of f (x) = ln x at x = 1 and use it to estimate ln1.12
L(x) =
ln1.12= �
To find the linear approximation of a function at a specific point, we use the formula for the tangent line to the function at that point.
The general formula for the tangent line to a function f(x) at x=a is:
L(x) = f(a) + f'(a)(x-a)
For the given function f(x) = ln(x) and the point x=1, we need to find the values of f(1) and f'(1).
1. Finding f(1):
Plug x=1 into the function f(x) = ln(x)
f(1) = ln(1) = 0
2. Finding f'(1):
Differentiate the function f(x) = ln(x) with respect to x to find its derivative.
f'(x) = 1/x (by the derivative of ln(x) is 1/x)
Plug x=1 into f'(x) to find f'(1)
f'(1) = 1/1 = 1
Now we can substitute the values of f(1) = 0 and f'(1) = 1 into the formula for the tangent line:
L(x) = f(1) + f'(1)(x-1)
L(x) = 0 + 1(x-1)
L(x) = x-1
To estimate ln(1.12), we can substitute x=1.12 into the linear approximation L(x):
L(1.12) = 1.12 - 1
L(1.12) = 0.12
Therefore, the linear approximation of f(x) = ln(x) at x=1 is L(x) = x-1, and the estimate for ln(1.12) using this linear approximation is 0.12.