Question

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

Answer 1

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`