Use differentials to approximate the square root of 24

Answer 1

with f(x) = x^1/2, we have f(25) = 5 and, since f '(x) = 1/2x^1/2 we have f '(25) = 1/2*5 = 0.1

Therefore f(24) ≈ f(5) + (24 - 25)*f '(25) = 5 - 0.1 = 4.9