Estimate using Linearization

Question

Estimate sqrt 25.1 using linearization L(x) of f(x) = sqrt x at a = 25 (Round to three decimal places)

L(25.1) =

Then find the actual value of sqrt 25.1

Answer 1

Greetings! Lets solve this shall we?

Let the formula for the linearization yield L(x) = f(a) + f'(a)(x-a).

f(a) = f(25) = √25 = 5

f'(a) = f'(25) = (1/2√25) = (1/10)

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

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

= 5 + (1/10)x - 2.5

L(x) = (1/10)x + 2.5

Now lets approximate a = 25.1

Then, L(x) = f(25.1) + f'(25.1)(x-25.1)

= 5.00999002 + .099800598(x-25.1)

L(x) = .100x + 2.505

I hope this helped!