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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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 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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 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. Tags linearization, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Kovacs Applied Math 1 April 23rd, 2016 04:12 AM complicatemodulus Number Theory 16 July 20th, 2014 12:30 AM jaredbeach Calculus 2 December 15th, 2011 04:36 PM ckooiker Calculus 2 November 3rd, 2011 09:08 PM rgarcia128 Calculus 2 October 17th, 2011 06:49 AM

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