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Estimate sqrt(99.8) using differentials (or equivalently, a linear approximation).

1 Answer

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The formula for a linearized function about a point x=a is :


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

Take a = 100


f(x)= √(x) \\f'(x)= (√(x) )'= (1)/(2 √(x) ) \\ \\f(100)= √(100)=10 \\ f'(100)= (1)/(2 √(100) )= (1)/(2 * 10 )= (1)/(20 ) \\ \\L(99.8)=10 + (1)/(20 )(99.8-100)=10- (0.2)/(20)= 9.99
User Lars Schillingmann
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