32.05^(4/5)
1 Answer

Put f(x) = x^(4/5) and do a linearization about the point x_0 = 32. Note that 32.05 = x_0 + Δx where Δx = 0.05.
f(x_0 + Δx) ≈ f(x_0) + f ‘(x_0)Δx
f ‘ (x) = (4/5) x^(1/5) ==> f ‘ (x_0) = (4/5)(32)^(1/5) = (4/5)/2 = 2/5.
(32.05)^(4/5) ≈ (32)^(4/5) + (2/5)(0.05) = 16 + 1/50 = 16.02.
A calculator generated decimal approximation to the value is 16.019997—very close!