Use differentials to approximate the value of the expression. compare your answer with that of a calculator. (round your answers to four decimal places.) a) sqrt(24.4) b) (1.99)^3

  a) approximate: 4.9400; calculator: 4.9396

  b) approximate: 7.8800; calculator: 7.8806

Step-by-step explanation:

The linear approximation to a function f(x) can be written as ...

  f(x+∆x) ≈ f(x) +(∆x)f'(x)

a) f(x) = √x; x = 25; ∆x = -0.6; f'(x) = 1/(2√x)

  f(x) = 5; f'(x) = 1/(2·5) = 1/10

  f(24.4) ≈ 5 + (-0.6)(1/10) = 4.9400 . . . approximation

  f(24.4) = √24.4 ≈ 4.9396 . . .  by calculator

The approximation is about 0.0074% high, a negligible amount for many applications.

b) f(x) = x^3; x = 2; ∆x = -0.01; f'(x) = 3x^2

  f(2) = 8; f'(2) = 12

  f(1.99) ≈ 8 + (-0.01)(12) = 7.8800 . . . approximation

  f(1.99) = 7.8806 . . . by calculator

The approximation is about 0.0076% low, a negligible amount for many applications.

the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24. hope this ! mark brainliest! you v much! : )