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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.

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01:33

Amrita Bhasin

04:28

Linda Hand

04:23

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 10

Linear Approximation and Differentials

Derivatives

Differentiation

Baro M.

October 15, 2020

Why didnot you illustrate by graphing ?

Missouri State University

Campbell University

Idaho State University

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

44:57

In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.

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well we find in the linear approximation to the function square root of one minus x. At equal zero. And we are you going to use that to approximate the numbers Square root of 0.9. and squared of 0.99 illustrate the situation by graphene F. And the tangent line. So we know that the function Can be approximated by its tangent line at zero. And that is given by the Taylor expression effort zero plus the derivative of F 0/1 factorial times X minus zero power one. Which is uh is simplified to have at zero plus F derivative at zero because one factorial is one and X two the one power. So this is the linear approximation to F. Yeah. At equal zero Or Taylor Polynomial of Degree one. And we recognize this is in fact the tangent line of F0 because we have the derivative of serious city slope of the line. And the independent values that fat zero. And the factor of the slope of the line is x minus zero. So it's just the tender line of F at zero. And we use it as a linear approximation to the function at equal. Super. So we've got to find the first derivative of path. This is a relative respect to eggs of the expression want square with 11 sex which can be written us one minus X to the one half. And Derivative of that is 1/2 times one minus X. To the one half minus one times is in the shame rule The relative of one -X. That's equal to one half times one minus X. To the negative one half Times -1. Because the derivative of the expression is 1 -1. And that's equal to negative one over two square root of 1 -6. So the first derivative of f is nearly 1/2 times the square root of 196. So that derivative at zero. He's not the 1/2. Because you will put zero here we get discovered one minus the risk or one which is one. So we get that. And we also need to F at zero. It's easy because that's that's square root of one minus zero is square one which is one. So is this two values too Right? The linear approximation to F at zero. That is F is approximated by the line. If it's zero is 1 plus the relative of F. Zero which is negative one half time sex or one minus X. Over to This is a straight line with slope negative one houses. Um align with doug slow And now we are going to use this to approximate the scourge of 3.9. And the skirts of 3.99. So for that we Look at this we know that the function is squared of 1 -6 and that is approximated by steve Line. The change in line at zero and Mhm Disease. To approximate the spirit of 0.9. We got to write this expression inside the squirrel As 1- Another value. That is that is equal to discover the one minus some value. And that's why you got to be 0.1 because one minus 3.13 point nine. And written this way we identify that this square, it is the image through the function F A Syrup .1. And that Is approximated by the line at 0.1. That is 1 -0.1 over two. I remember one is 1/10. So this is 1/20 And this is 19/20 Which is equal to zero 95. Now with the same force these words of 0.99. We know difficult to discourage 1 -0.01. So this image through f of 0.01 which is approximated by 1 -0.01 Over to that is 1 -1 over 200. Because 0.01 is one over 100. This is 200 point of Swan is 199. Over 200. And this value is 0.995. So the square root of 0.9, He separates ultimately equal to 0.95 and square to 0.99, separate, ultimately equal to 0.95. And we can see that all the graphs of the functions we are going to display at the moment. Mhm. Mhm. So we have here uh the graph of the function first on the interval and zero ones. We can see here. And the red line is the tangent line at zero and the blue line is the function. So I put it here. Sorry. So this red line is the tender line which is the linear approximations to function at equal serum. And the function itself is this blue line here. And I just draw this to notice we have A good approximation near the point of expansion of tells D. Taylor polynomial of degree one that is zero Closer to zero. We have a good approximation and we go away from zero. The approximation is not so very good. So we have used uh As point of revelation 0.1 and 0.01 which are actively close to zero. So we have we get to having good approximations to the square roots. We are looking forward So we gotta do is some zooming here around zero, just between Syrian 0.1. Because there are these this interval, We have the values, we are evaluating the function 0.1 And 0.01. So we have this soon here. What? Yeah, okay, so we have here the function between zero and 0.1 And there we are evaluating at 0.1 At 0.01. So we are evaluating first at this at this point. Sorry here at this point and as we can see the value we get from The line which is the red line in this graph, we see that we have 0.95 which is just we calculated here above Yeah. And with this values your .95 we are close to the real value which will be given by looking at the recent a line through this point here Should be a value from here or less. So we are relatively close to the value 0.95 To the real value of the function at 0.1. But when we evaluated 0.01 which is the expression, we can see that we are in a region where the functional line are indistinguishable so they are the same practical. So the value we are getting here 0.995 must be a very good approximation To the real value of the square roots of 0.99. So that's the situation here. So when we are getting closer to zero, the values uh of approximation given by the linear approximation to the function are relatively acceptable or good. But the situation, it's not the same or the situation is getting worse. When we go away from where we are getting far from the value of um where we found the tangent line so serious in fact is tangent line, it is another degree of confidence. That is, it's not a line but well number two or three, we will have a very good approximation. A little bit, a little uh, an interval a little bit greater than that. We do with italian lines. And To be the tangent lines, a good approximation at 0.01. For example, here we have, we see a gap already but maybe we can except the value forced to open one Even if we are not so close, but when we are using attention line for values like for example, 0.8 Or a 0.9, 0.5 etc. These values here we see a gap is very big. So no use in using convenient to use the tangent line at zero to approximate these values. We have always these um alternative to develop the change in line at another point. So if we are going to avoid an other point, we must use Nazi era, but a different point of evaluation of the tangent or development of the tangent line. So these are the approximations and this is the graphical situation to the approximations

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