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

1 Answer
Jul 31, 2015

Answer:

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

Explanation:

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)