# Use linear approximation to estimate the number sqrt(9.05)

QUESTION POSTED AT 01/06/2020 - 04:48 PM

QUESTION POSTED AT 01/06/2020 - 04:48 PM

Linear approximation is used when the value needed is close to a known value.

Here, we know sqrt(9)=3.

Let

f(x)=sqrt(x)

f'(x)=1/(2sqrt(x))

Apply linear approximation

f(9.05)=f(9)+f'(x)dx

=sqrt(9)+0.05/(2sqrt(9))

=3+.05/(6)

=3.008333

exact value is 3.008322, estimate is off by .0004%

Here, we know sqrt(9)=3.

Let

f(x)=sqrt(x)

f'(x)=1/(2sqrt(x))

Apply linear approximation

f(9.05)=f(9)+f'(x)dx

=sqrt(9)+0.05/(2sqrt(9))

=3+.05/(6)

=3.008333

exact value is 3.008322, estimate is off by .0004%

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