Find the linear approximation of f(x)=ln x at x=1

where L(x)=

y = ln x

m = dy/dx = 1/x
at x = 1, m = 1/1 = 1
so
y = x + b
at x = 1, y = ln 1 = 0
so
0 = 1*1 + b
so
b = -1
so
y = x - 1

To find the linear approximation of a function f(x) at a specific value x=a, you need to use a tangent line to approximate the function near that point. The equation of a line can be written as y = mx + b, where m is the slope of the line and b is the y-intercept.

In this case, we want to find the linear approximation of f(x) = ln(x) at x = 1. To do this, we need to find the slope of the tangent line at x = 1.

The slope of the tangent line to a function at a specific point can be found using the derivative of the function. The derivative of f(x) = ln(x) is given by f'(x) = 1/x.

Now, we can find the slope at x = 1:
f'(1) = 1/1 = 1

So, the slope of the line (m) will be 1.

To find the y-intercept (b), we can use the point-slope form equation y - y1 = m(x - x1), where (x1, y1) is a point on the line (in our case, x1 = 1 and y1 = ln(1)).

Since ln(1) = 0, we have:
y - 0 = 1(x - 1)
y = x - 1

Therefore, the linear approximation of f(x) = ln(x) at x = 1 is given by L(x) = x - 1.