Question #c3851

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Answer 1 · 2017-05-14T18:25:34

$f (0.4983) \approx (\frac{π}{6} - 0.1963) r a d i a n .$ $f(0.4983)~~(pi/6-0.1963) radian.$

$f(x)=arc sinx rArr f'(x)=1/sqrt(1-x^2).$

Knowing that,

$f(a+h)~~f(a)+hf'(a)," we subst., a=0.5, and, h=-0.0017,"$ & get,

$f(0.5-0.0017)~~f(0.5)-0.0017f'(0.5), or,$

$f(0.4983)~~f(0.5)-0.0017f'(0.5).$

$:. arc sin(0.4983)~~arc sin0.5-0.17/sqrt(1-0.5^2)$

$=pi/6-0.17/sqrt0.75=pi/6-0.17/(0.5sqrt3)~~pi/6-0.17/0.866$

$rArr f(0.4983)~~(pi/6-0.1963) radian.$