How do you find the linearization of #f(x)=lnx # at x=8?

Question

Isaiah Allen · Accepted Answer

From simple geometry, plus some complicated things to do with general smoothness of the curve about #x = 8#, we can say for "small" #epsilon# that:

#f(8 + epsilon) approx f(8) + epsilon f'(8)#

And because #f(x) = ln x# then #f'(x) = 1/x#; and we can say that:

#f(8 + epsilon) approx ln 8 + epsilon 1/8 #

We can test this in a calculator for #f(8.1)#

Actual Value: #ln 8.1 = 2.0919#

From Linearisation: #ln 8 + 0.1 * 1/8 = 2.0919#

!!

For #f(8.9)#

Actual Value: #ln 8.9 = 2.186#

From Linearisation: #ln 8 + 0.9 * 1/8 = 2.192#

:(

For #f(20)#

Actual Value: #ln 20 approx 3#

From Linearisation: #ln 8 + 12 * 1/8 approx 3.6#

:((

Aurora Alvarado · Answer

undefined To find the linearization of $ f(x) = \ln(x) $ at $ x = 8 $, follow these steps:

1. Calculate the first derivative of $ f(x) $ with respect to $ x $, denoted as $ f'(x) $.
$$ f'(x) = \frac{1}{x} $$

2. Evaluate $ f'(x) $ at $ x = 8 $.
$$ f'(8) = \frac{1}{8} $$

3. Find the value of $ f(8) $, which is simply $ \ln(8) $.

4. Use the point-slope form of a linear equation to construct the linearization:
$$ L(x) = f(a) + f'(a)(x - a) $$
where $ a $ is the point of tangency, in this case, $ a = 8 $.

5. Substitute the values of $ f(a) $, $ f'(a) $, and $ a $ into the equation to get the linearization. Thus, the linearization of $ f(x) = \ln(x) $ at $ x = 8 $ is:
$$ L(x) = \ln(8) + \frac{1}{8}(x - 8) $$