# How do you find the linear approximation of (1.999)^4 ?

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35
Aug 17, 2014

You can use the tangent line approximation to create a linear function that gives a really close answer.

Let's put $f \left(x\right) = {x}^{4} ,$ we want $f \left(1.999\right)$ so use x= 1.999 and the nearby point of tangency a = 2. We'll need $f ' \left(x\right) = 4 {x}^{3}$ too.

The linear approximation we want (see my other answer) is

$f \left(x\right) \approx f \left(a\right) + f ' \left(a\right) \left(x - a\right)$

$f \left(1.999\right) \approx f \left(2\right) + f ' \left(2\right) \left(1.999 - 2\right)$

$\approx {2}^{4} + 4 \cdot {2}^{3} \cdot \left(- 0.001\right) = 16 - 0.032 = 15.968$

You can compare to the actual exact result of
${1.999}^{4} = 15.968023992001 ,$so we came pretty close!

Bonus insight: The error depends on higher derivatives and can be predicted in advance! \ dansmath strikes again, approximately! /

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