How do you find the linear approximation of #f(x)=ln x# at x=1 and use it to estimate ln 1.42?

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Answer 1 · 2017-11-05T09:28:55

The first estimate is #=0.42#

The method is

#f(x+h)=f(x)+hf'(x)#

#f(x)=lnx#

#f'(x)=1/x#

#x=1#

#h=0.42#

Therefore,

#f(1.42)=ln(1.42)=f(1)+0.42f'(1)=0+.42=0.42#

You can also apply the Taylor 's series for #ln(1+x)#

#ln(1+x)=x-x^2/2+x^3/3+o(x^3)#

#ln(1.42)=0.42-0.42^2/2+0.42^3/3=0.356#

The value of #ln1.42=0.351#

The error is #=0.351-.0356=-0.005#