How do you use linear approximation to estimate #root3( 64.1)#?

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Answer 1 · 2016-07-20T08:01:08

#root(3)64.1=4.002083#.

For a fun. #f, f(x+h)~=f(x)+hf'(x).................(1)#.

Consider the fun. #f(x)=root(3)x=x^(1/3) rArr f'(x)=1/3*x^(-2/3)#

We take #x=64, and, h=0.1# in #(1)#. Therefore,

#f(64+0.1)~=f(64)+0.1f'(64)#

#:.(64.1)^(1/3)~=64^(1/3)+0.1*1/3*64^(-2/3)=4+0.1*1/3*4^(-2)#

#=4+1/10*1/3*1/16=4+1/480~=4+0.002083=4.002083#.

Thus, #root(3)64.1=4.002083#.