Consider the function f(x) = x ³. Since 2³ = 8 is a perfect cube that is pretty close to x = 1.999, take the linear approximation to f(x) at x = 2. It is
f(x) ≈ L(x) = f (2) + f ' (2) (x - 2)
Compute the derivative:
f(x) = x ³ → f ' (x) = 3x ² → f ' (2) = 12
Then
L(x) = 8 + 12 (x - 2) = 12x - 16
Now approximate 1.999³ :
f (1.999) ≈ L (1.999) = 12•1.999 - 16 = 7.988
(Compare this to the actual value, 7.98801.)
