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Find the linear approximation of the function $f(x)=\sqrt{1-x}$ at $a=0$ and use it to approximate the numbers $\sqrt{0.9}$ and $\sqrt{0.99}$ . Illustrate by graphing $f$ and the tangent line.

$L(x)=1-\frac{1}{2}(x-a)=1-\frac{1}{2} x$$L(0.01)=1-\frac{1}{2} 0.01=0.95$$L(0.01)=1-\frac{1}{2} 0.001=0.995$

Calculus 1 / AB

Chapter 3

Derivatives

Section 8

Linear Approximations and Taylor Polynomials

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find a linear approximation of the function. F of X equals the square root of one minus X at X equals zero and use it to approximate square to a 00.9 square, 2.99 So if you look at the diagram here, the red function is the square toe, one minus x. And then the blue line is the approximation, the linear approximation that we're going to get here in just a minute and you can see that near zero. There are almost exactly the same thing. All right, So to find a linear approximation, you have to have a point. So they told us to use X equals zero, and so f of zero equals the square. No. One minus zero, which is one. So our 0.201 and then we need the slope. And to find the slope, we have to take the derivative. It's one half one minus X to the minus one half times. Don't forget the derivative of the inside, which is negative one. So you get negative 1/2 times the square toe, one minus six. Okay, so that's the slope of any tangent line to that red curve, but we want the specific one, the blue one. So we need a plug in zero to find out what it ISS so am equals f prime of zero, which is minus 1/2 times the square root of one minus zero. So minus one half, so is our point. Here's our slope, or we have to do is write the equation of the line so minus one half equals Y minus one over X minus zero Cross multiply two y minus two equals negative X two y equals negative X plus two. So why equals negative one half X plus one? So that's our linear approximation. If Okay, so what it means is to calculate I square root that near the square root of one instead of calculating it or getting out your calculator, you can just use this nice, easy polynomial equation. So, for example, if we wanted to find the square 2.9, we said it equal to the square root of one minus X, so we can figure out what X to use. Well, X would be 10.1 there. So then the square 2.9 is approximately equal to minus one half times 0.1 plus one from this equation right here. Okay, One half of 10.1 is 0.5 So minus 0.5 plus one 0.95 So if you get out your calculator and you calculate that square 2.9 is 0.948 something something, something. So look at what a good approximation I got without ever having to take the square root of anything. And then 0.99 this one we're gonna use X equals point. No one can. The closer you get to zero, the closer X is 20 The better the approximation is gonna be. So 0.99 is approximately square 2.9 approximately equal to minus one half times point Oh, one plus one. So that's minus 0.5 which is 0.995 and the real square root of 0.99 according to my calculator or something like 0.9948 something something, something. So if you go back and look at the graph, you can see that as long as you pick a number in between about here and here, you're going to get a really good answer. But if you get too far away from zero like here, you're trying to get this. But your approximation is going to give you that instead. That's because you're too far away from the point that you wrote the approximation near. All right, that's how you make linear approximations.

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