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Use linearization to approximate the given quantity. In each case determine whether the result is too large or too small.(a) $\sqrt{16.01}$(b) $\sqrt{15.98}$

(a) 4.00125 large(b) 3.9975 large

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 6

Linearization and Differentials

Derivatives

Missouri State University

Campbell University

Harvey Mudd College

Boston College

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.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

04:04

Use linearization to appro…

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Use the linearization $(1+…

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05:20

Let $f(x)=\sqrt{4+5 x},$ l…

in this question. We are to approximate Beilin arr ization to function the number 16.01 and 15.98. No We know that the roots of these numbers are near the root of 16. So we're gonna use uh 16 as our it's not and our function will be to go to the root of X. 32 facts which is also equal to X. to the poor half. So we know that for a linearize function equation we need F. Okay plus have prime of a. Well played by X minus A. Now. Uh huh. We need a prime of its which is the part comes down gives us half X. And we subtract -1 in the power we get minus half. So this is F. Prime of X. Which can be simplified to one over to root of X. Now substituting we need um F. A. We just substitute our a. Which is our .16 into the equation. FFX now 1316. We get root of 16 which is equal to four F. Prime of a. We substitute 16 or so into the F. Prime of X equation. We get 1/8. Now calculating our Alex, We get our Alex s. four plus if F. Prime of A. Is um He's 1/2 is one of 8. 1/8 X- What is our 8:16. Simplifying this, we get to plants 1/8 x. Now we know that we want to linearize. Um We want to find the root of 16.01. So we Plugging 16.01 Jar linearize function equation. Then we we find The route approximation by common um multiplication and division here to be to to be four 00125. Yeah let me put that nice little bit nicer to be four 0.0 125. Now if you plug into a calculator you want to just find the root of 16.01, you will find that it is. Uh this result is actually larger than what you can expect from that from the calculator. Next we're gonna um just Figure out also the root of 15.98 For 15.98 we will use the same function And the same derivatives. So and the same x. note which is 16. So we are just coming back to our equation. L affects Is he co two two Plus 1/8 X. Now here we know that our X. Since you want to find the root of 15.98 you know Alex you're going to substitute 15.98 here you get two plus 15.98 over eight. Now if you calculate this uh normal calculation you're gonna get the root as three point 9975. And also this is larger than their Uh than the number you would find from plugging in 15.98 into a calculator. So you're done

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