Question #91801

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Answer 1 · 2014-11-18T11:03:28

Linearization of f(x) at a

#L(x)=f(a)+f'(a)(x-a)#

Let #f(x)=ln x#. #=> f(1)=ln(1)=0#

#f'(x)=1/x => f'(1)=1/{(1)}=1#

So, the linearization #L(x)# of #f(x)# at #1# is

#L(x)=0+1(x-1)=x-1#

Hence, we can approximate

#ln(1.003) approx L(1.003)=(1.003)-1=0.003#

I hope that this was helpful.