Expert: Paul Klarreich - 2/27/2006

Question
Find the linear approximation of the function
g(x)= (1+x)^(1/3) at a=0 and use it to approximate the numbers ('0'.95)^(1/3) and (1.1)^(1/3).

Answer
chevonne Asks in Category Calculus:
Subject:  linear approximations
Question:  Find the linear approximation of the function
g(x)= (1+x)^(1/3) at a=0 and use it to approximate the numbers

('0'.95)^(1/3) and (1.1)^(1/3).
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When you do a linearization (lin. approx.) you make the following assumptions:

1. y1 = f(x1),  and y2 = f(x2), and you want the value of y2 for a given x2.

2. y2 = y1 + Dy    [Dy = 'delta-y'; I can't make a delta.]

3. Dy/Dx ~~ dy/dx  [~~ means 'approximately equal to']

So you choose  x1 that is near x2, but for which the calculation of f(x1) is MUCH EASIER.  

In this example, y = f(x) = (1+x)^1/3,   you choose x = 0 because (1)^1/3 is easy to compute -- it is equal to 1.

So  dy/dx = 1/3(1+x)^(-2/3)

At x = 0,  dy/dx = 1/3, and we will use  Dy = Dx/3

For your  (0.95)^1/3, take:

x1 = 0
y1 = 1
Dx = -0.05
Dy = -0.01667
y2 = y1 + Dy = 1 - 0.01667 = 0.98333..

For your  (1.1)^1/3, take:

x1 = 0
y1 = 1
Dx = 0.1
Dy = 0.03333..
y2 = y1 + Dy = 1 + 0.03333 = 1.03333..