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प्रश्न
Find the approximate value of ` sqrt8.95 `
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उत्तर
`Let f(x)=sqrtx`
`f(a+h)=sqrt(a+h)`
we choose a=9 and h=-0.05
Then `sqrt8.95=f(a+h) `
But `, f(a+h)~~f(a)+hf'(a)`
`sqrt8.95~~f(a)+hf'(a)...........(1)`
Now ` f(a)=sqrta=sqrt9 and h=-0.05`
we have `f(X)=sqrtx `
`f'(x)=1/(2sqrtx)`
`f'(a)=f'(9)=1/(2sqrt9)=1/(2xx3)=1/6`
from(1) ` sqrt(8.95)~~3+(-0.05)(1/6)`
=3-0.0083=2.9917
`=> sqrt8.95 ~~2.9917`
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