Find the linear approximation of f(x)=lnx at x=1 and use it to estimate ln1.12. L(x)=?ln1.12=?

Expert Answers

An illustration of the letter 'A' in a speech bubbles

We are given `f(x)=ln x` and we are interested in approximating the value of `f(x)` near `x=1` :

(1) `f(1)=0`

(2) `f'(x)=1/x,x>0`

(3) Thus the slope of the line tangent to the curve at (1,0) is `f'(1)=1`

(4) The equation of the line tangent to `f(x)` at (1,0) is:

`y-0=1(x-1) "or" y=x-1`

For `x` close to 1, the output for this line is approximately the same as the output for `f(x)` . If `L(x)=x-1` then `L(x)~~f(x)` for all x close to 1-- the closer to 1 the better the approximation.

(5) So `L(x)=x-1` and `f(1.12)~~L(1.12)=.12`

** The actual value for `ln(1.12)~~0.113328685307`

Approved by eNotes Editorial Team

We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

  • 30,000+ book summaries
  • 20% study tools discount
  • Ad-free content
  • PDF downloads
  • 300,000+ answers
  • 5-star customer support
Start your 48-Hour Free Trial