My Math Forum Linearization problem

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 February 22nd, 2019, 02:28 PM #1 Newbie   Joined: Feb 2019 From: East Lansing, MI Posts: 3 Thanks: 0 Linearization problem https://ibb.co/BcY1dxJ Find the best function f(x) and value a so that the linearization of f(x) at x=a can be used to estimate √16.2. Find the linearization of f(x) at a. Use the linear approximation from (b) to estimate √16.2. I think I can figure out the last two, but I'm not sure how to get the function. Last edited by skipjack; February 22nd, 2019 at 11:14 PM.
 February 22nd, 2019, 02:56 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,468 Thanks: 1342 Taking the first 2 terms of the Taylor series of the square root function about $x_0$ we have $\sqrt{x} \approx \sqrt{x_0} + \dfrac{x-x_0}{2\sqrt{x_0}}$ hopefully clearly to find $\sqrt{16.2}$ we'll choose $x_0 = 16$ as it is the closest perfect square. We then end up with $\sqrt{16.2} \approx \sqrt{16} + \dfrac{16.2-16}{2 \cdot 4} =$ $4 + 0.025 = 4.025$ which is a pretty good approximation.
February 22nd, 2019, 03:20 PM   #3
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Quote:
 Originally Posted by romsek Taking the first 2 terms of the Taylor series of the square root function about $x_0$ we have $\sqrt{x} \approx \sqrt{x_0} + \dfrac{x-x_0}{2\sqrt{x_0}}$ hopefully clearly to find $\sqrt{16.2}$ we'll choose $x_0 = 16$ as it is the closest perfect square. We then end up with $\sqrt{16.2} \approx \sqrt{16} + \dfrac{16.2-16}{2 \cdot 4} =$ $4 + 0.025 = 4.025$ which is a pretty good approximation.
It didn't show up as correct for me. :/ It can't be decimals but even when I converted it to a fraction it said it's wrong. Sorry

February 22nd, 2019, 03:58 PM   #4
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Quote:
 Originally Posted by apertxre https://ibb.co/BcY1dxJ Find the best function f(x) and value a so that the linearization of f(x) at x=a can be used to estimate √16.2. Find the linearization of f(x) at a. Use the linear approximation from (b) to estimate √16.2. I think i can figure out the last two, but I'm not sure how to get the function.
This is a terrible question. If you fix a value at which to linearize, then the linearization is the best approximation choice. But if you don't fix the value, then the question is absurd. I would guess they want you to linearize at $a = 16$ but this certainly won't be the best choice by any means.

Here is a linear approximation which is better: $y = \sqrt{16.2}$.

Here is another one which is less trivial: $y =\sqrt{16.2} + \frac{1}{2 \sqrt{16.2}}(x - \sqrt{16.2})$.

Whoever assigned this question should be slapped.

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