Find the linearization l(x) of the function at a. f(x) = x^1/2 , a = 25.

Solution:

Given, the function f(x) = \((x)^{\frac{1}{2}}\)

We have to find the linearization L(x) of the function at a = 25.

Using the formula,

L(x) = f(a) + f’(a)(x - a)

Now,

f(x) = \((x)^{\frac{1}{2}}\)

f(a) = f(25) = √25 

f(a) = 5

f’(x) = (√(1 - x))’

= (1/2)(1/√(x))

f’(a) = f’(25) = (1/2)(1/√(25))

= (1/2)(1/5)

= 1/10

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

L(x) = 5 + (1/10) (x - 25)

L(x) = 5 + (1/10)x - (25/10)

L(x) = 5 + x/10 - 5/2

L(x) = (10-5)/2 + x/10

L(x) = 5/2 + x/10

Therefore, the linearization of f(x) = \((x)^{\frac{1}{2}}\) at a = 25 is L(x) = 5/2 + x/10.

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Find the linearization l(x) of the function at a. f(x) = x^1/2 , a = 25.

Summary:

The linearization of the function f(x) = \((x)^{\frac{1}{2}}\) at a = 25 is L(x) = 5/2 + x/10.