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

$\sqrt{1-x} \approx 1-\frac{1}{2} x ; \sqrt{0.9} \approx 0.95 ; \sqrt{0.99} \approx 0.995$

04:32

Jack C.

Calculus 1 / AB

Chapter 2

DERIVATIVES

Section 8

Linear Approximations and Differentials

Derivatives

Differentiation

Applications of the Derivative

Campbell University

Harvey Mudd College

University of Nottingham

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find the linear approximation for the function. F of X equals the square root of one minus X at X equals zero. Use it to approximate the square 2.9 and the square 2.99 So even though x zero were ableto calculate the square roots near one because we're subtracting, um, numbers from one. Okay, but if you get very far away from one, then it's not gonna work. Okay, So, ffx, that's this red graph right here, the square root of one minus X. And now we're going to write the linear approximation, which is the blue line. So first, we need to find a point on the line. Okay, Can the point they want us to use is where x zero and you plug zero in and you get why is one cool and then we need the slope and find the slope. We take the derivative of prime of X. It's one half one minus X to the minus one half times minus one. That minus one, came from the derivative of the inside. So it's minus 1/2 times the square toe, one minus x. So that's the slope of any tangent to that curve, and we want the specific one where x zero. So we plug in X equals zero. And now it's the slope. Am equals f prime of zero, which is minus 1/2. And now I'm going to write the equation of the line. Yeah, slow equals. Why? Minus y over X minus X cross. Multiply solve for why, um two y equals minus X plus two. Or why equals minus one half X plus one. Okay, so let's see if that's the line I have drawn. Hopefully it is Y intercept is one. A slope is negative. One half. All right, so we want to calculate squared of 0.9. So we started equal to the square root of one minus X so we can figure out what X is so square. Both sides 0.9 equals one minus x, so X must be equal 2.1. Oh, snap 0.9 point one. So then here's our approximation Then the square root of 0.9 is approximately equal to minus one half times 10.1, which is 1/10 plus one. This is minus 1/20 plus one. So 19 20th, which is 0.95 So I'm saying the square root of 0.9 is approximately 0.95 So I got my calculator out. I see what it is. 0.9 square root, right? Actual value 0.9 0.94868 blah, blah blah. Okay, here's the beauty of this look. I never had to take the square root. I didn't have to get out my calculator. I could get a good approximation to the square 2.9 without even doing hardly any calculations. That's the beauty of the linear approximation. Now let's use it to 5.99 square root a 0.99 So this time X is equal 2.1 So then 0.99 is approximately equal to minus one half times point No. One, which is one over 100 plus one so minus one over 200 plus one. That's 199 over 200. Okay, yeah, so that's 99.5 over 100. So that's 0.995 So I'm saying the square root of 0.99 is 0.995 approximately. Let's see what calculator says. 0.99 square roots, according to my calculator is 0.994987 more stuff. Okay, It's a very good approximation to the square root of this. Okay, now, if you look back at the picture here as long as you're taking the square root of numbers that fall within here, then, um, this line is a good approximation. But if, for example, you wanted to use X equals one, it's not a good approximation. Or over here, it's not a good approximation. You need to stay where the line and the curve or close to each other.

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