Use a linear approximation (or differentials) to estimate the given number [Solved]

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Put f(x) = x^(4/5) and do a linearization about the point x_0 = 32. Note that 32.05 = x_0 x where x = 0.05.
f(x_0 x) f(x_0) f (x_0)x
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 16.019997very close!

Put f(x) = x^(4/5) and do a linearization about the point x_0 = 32. Note that 32.05 = x_0 x where x = 0.05.
f(x_0 x) f(x_0) f (x_0)x
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 16.019997very close!

This post is last updated on hrtanswers.com at Date : 1st of September – 2022

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