How do you find the linearization of #f(x) = 2 ln(x)# at the point a=1?

Question

Emily Ammerman · Accepted Answer

The linearization is the equation of the tangent line (often written in a particular form).

#f(x) = 2ln(x)# at #x=1#, f(1) = 0# #f'(x) = 2/x#, so #f'(1) = 2# We get the linearization (and the tangent line) using #y - f(a) = f'(a)(x-a)# In this case #y = 0 + 2 (x-1)# Rewrite to taste (or the taste of the person grading your work.)

Evelyn Agard · Answer

undefined To find the linearization of $ f(x) = 2 \ln(x) $ at the point $ a = 1 $, we follow these steps:

1. Find the first derivative of $ f(x) $, denoted as $ f'(x) $.
2. Evaluate $ f'(1) $ to find the slope of the tangent line at $ x = 1 $.
3. Use the point-slope form of a line, $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) $ is the point $ (1, f(1)) $ and $ m $ is the slope we found earlier, to write the equation of the tangent line.

Let's go through these steps:

1. The derivative of $ f(x) = 2 \ln(x) $ is $ f'(x) = \frac{2}{x} $.
2. Evaluate $ f'(1) $: $ f'(1) = \frac{2}{1} = 2 $.
3. Using the point-slope form with $ (x_1, y_1) = (1, 2 \ln(1)) = (1, 0) $ and $ m = 2 $, the equation of the tangent line is $ y - 0 = 2(x - 1) $.

Simplify the equation to get the linearization of $ f(x) $ at $ a = 1 $: $ y = 2x - 2 $.