How do you find the linear approximation of #f(x)=ln(x)# at #x=1# ?

Question

Isaiah Allen · Accepted Answer

Here is the big key: The linear approximation of #f# at #a# is the tangent line at #a#.

The linear approximation of #f(x)# at #x=a# is given by: #L(x) = f(a) + f'(a) (x-a)# The equation of the tangent line to the graph of #f# at #(a, f(a))# is the equation of a line through #(a, f(a))# whose slope is #f'(a)# . In point-slope form, the line is: #y-f(a) = f'(a)(x-a)# So we can write the tangent line as: #y = f(a) +f'(a)(x-a)# I struggled mightily with linear approximation for a week or two before this hit home and I realized that linear approximation is just a particular way of thinking about, writing, and using the tangent line. (My teacher had been saying it, I just didn't see it.) In this problem For #f(x) = lnx#, we have #f'(x) = 1/x#. Therefore, #f'(1) = 1/1 = 1# We also not that #f(1) = ln(1) =0#. The linear approximation is the line: #y-0 = 1(x-1)# Or, simply #y = x-1# If you have a calculator of tables for #ln# you can quickly see that #{: (x," calculator" ln(x), " approx by "x-1),(1.05," "" "0.04879," "" "0.05),(1.01," "" "0.00995," "" "0.01),(0.997," "-.0.003005," "-0.003) :}# Note, however that #ln(2) ~~ 0.6931# While the linear approximation gives #2-1 = 1# (not a very good approximation for many purposes)

Scarlett Allison · Answer

undefined To find the linear approximation of $ f(x) = \ln(x) $ at $ x = 1 $, we first find the first derivative of $ f(x) $, which is $ f'(x) = \frac{1}{x} $. Then, we evaluate $ f'(1) $ to find the slope of the tangent line at $ x = 1 $. Since $ f'(x) = \frac{1}{x} $, $ f'(1) = 1 $.

Next, we use the point-slope form of a linear equation to write the equation of the tangent line:

$$ y - f(1) = f'(1)(x - 1) $$

Since $ f(1) = \ln(1) = 0 $, we have:

$$ y - 0 = 1 \cdot (x - 1) $$

This simplifies to:

$$ y = x - 1 $$

Therefore, the linear approximation of $ f(x) = \ln(x) $ at $ x = 1 $ is $ y = x - 1 $.