Determine the values of x for which the linear approximation is accurate to within 0.1. f(x) = 1/(1 + 2x)⁴

Solution:

Given, the function f(x) = 1/(1 + 2x)⁴

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) = 1/(1 + 2x)⁴

f(a) = f(0.1) = 0.4822530

To find f’(x),

Let u = (1 + 2x)

So, (1 + 2x)⁴ = u⁴

On differentiating,

By power rule, u⁴ = 4u³

Applying chain rule,

d(2x + 1)/dx = 2

So, 2(4u³) = 8(2x + 1)³

f’(x) = -8/(2x + 1)⁵

Now, f’(a) = f’(0.1) = -8/(2(0.1) + 1)⁵

f’(a) = -3.2150

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

L(x) = 0.4822530 + (-3.2150)(x - 0.1)

L(x) = 0.4822530 - 3.2150x + 0.3215

L (x) = 0.803753 - 3.2150x

Therefore, the linearization of f(x) = 1/(1 + 2x)⁴ at a = 0 is L(x) = 0.803753 - 3.2150x.

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Determine the values of x for which the linear approximation is accurate to within 0.1. f(x) = 1/(1 + 2x)⁴

Summary:

The linearization of the function f(x) = 1/(1 + 2x)⁴ at a = 0 is L(x) = 0.803753 - 3.2150x.