ANSWER

Step-by-step explanation
We want to find the linear approximation of the function at x = 1:

First, we have to find the equation of the tangent line to the function at x = 1:

Now, find the derivative of the function at x = 1:

Now, find the equation of the line using the point-slope method:

Therefore, we have:

Hence, the linear approximation of f(x) at x = 1 is:

To find the estimate of ln(1.46), substitute 1.46 for x in the equation above and simplify:

That is the answer.