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

Question

Ezra Adams · Accepted Answer

You can use the tangent line approximation to create a linear function that gives a really close answer. Let's put #f(x) = x^4,# we want #f(1.999)# so use x= 1.999 and the nearby point of tangency a = 2. We'll need #f'(x)=4x^3# too. The linear approximation we want (see my other answer) is  #f(x) ~~ f(a) + f'(a)(x-a)# #f(1.999) ~~  f(2) + f'(2)(1.999-2)# #~~ 2^4 + 4*2^3*(-0.001) = 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! /

William Abdalla · Answer

undefined To find the linear approximation of $ (1.999)^4 $, we start by choosing a point close to $ 1.999 $ to use as the center of our approximation. Let's use $ x = 2 $ since it's close to $ 1.999 $. Then, we need to find the first derivative of the function $ f(x) = x^4 $ and evaluate it at $ x = 2 $, which gives us $ f'(x) = 4x^3 $, and $ f'(2) = 32 $.

Now, we use the linear approximation formula:

$$ L(x) = f(a) + f'(a)(x - a) $$

Substitute $ a = 2 $ and $ f'(a) = 32 $ into the formula:

$$ L(x) = f(2) + 32(x - 2) $$

Now, find $ f(2) = 2^4 = 16 $:

$$ L(x) = 16 + 32(x - 2) $$

This is the linear approximation of $ (1.999)^4 $.