Find the linear approximation of the function f(x) =√( 4 - x) at a = 0 and use it to approximate the numbers √3.9 and √3.99?

Solution:

Given, the function f(x) = √(4 - x)

We have to find the linearization L(x) of the function at a = 0.

Using the formula,

L(x) = f(a) + f’(a)(x - a)

Now,

f(x) = √4 - x

f(a) = f(0) = √4

f(a) = 2

f’(x) = -(1/2)(1/√(4 - x))

f’(a) = f’(0) = -(1/2)(1/√(4 - 0))

= -(1/2)(1/√4)

= -(1/2)(1/2)

f’(a) = -1/4

Substituting the values of f(a) and f’(a), the function becomes

L(x) = 2 + (-1/4)(x - 0)

Therefore, the linearization of f(x) = 2 - (1/4)x at a = 0 is L(x) = 2 - (1/4)x.

If x = 0.1, we get √3.9

L(0.1) = 2 - (1/4) 0.1 

= 2 - 0.025 

= 1.9750

If x = 0.01, we get √3.99

L(0.01) = 2 - (1/4)(0.01)

= 2 - 0.0025

= 1.9975

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Find the linear approximation of the function f(x) = 4 - x at a = 0 and use it to approximate the numbers √3.9 and √3.99?

Summary:

The linearization of the function f(x) = √4 - x at a = 0 is L(x) = 2 - (1/4)x. Using it to approximate the numbers √3.9 and √3.99 we get 1.9750 and 1.9975