A. Find the linear approximating polynomial for the...

a. Find the linear approximating polynomial for the following functions centered at the given point a. b. Find the quadratic approximating polynomial for the following functions centered at the given point $a$. c. Use the polynomials obtained in parts (a) and (b) to approximate the given quantity. $$f(x)=\frac{1}{x}, a=1 ; \text { approximate } \frac{1}{1.05}$$

a. Find the linear approximating polynomial for the following functions centered at the given point a.
b. Find the quadratic approximating polynomial for the following functions centered at the given point $a$.
c. Use the polynomials obtained in parts (a) and (b) to approximate the given quantity.
$$f(x)=\frac{1}{x}, a=1 ; \text { approximate } \frac{1}{1.05}$$

Key Concepts

Quadratic Approximation

The quadratic approximation is the second-degree Taylor polynomial, incorporating both the first and second derivatives of the function at the center. This polynomial not only accounts for the slope of the function but also its curvature, leading to a more precise approximation compared to the linear version, particularly when moving slightly further away from the central point.

Taylor Polynomial

A Taylor polynomial is an approximation of a function around a specific point using the sum of its derivatives evaluated at that point. It provides a polynomial of a chosen degree that closely models the behavior of the original function near the center. For any smooth function, the Taylor polynomial of degree n uses the function's derivatives up to the nth order, allowing for a systematic refinement of the approximation as higher-order terms are added.

Linear Approximation

The linear approximation is the first-degree Taylor polynomial, which uses only the value of the function and its first derivative at the point of interest. It essentially finds the tangent line to the function at that point and is most accurate for values very close to the center. This method is a basic tool in calculus for estimating the function's value when a small change occurs near the known point.

Transcript

00:01 So you have this function f of x equals 1 over x and we want to find the linear approximation.

00:13 Linear approximation.

00:16 So this is the same.

00:18 The linear approximation is the same as the taylor polynomial of degree 1.

00:28 So this tailor polynomial of degree 1, well centered at a equals 1.

00:39 This is the scalar polynomial of degree 1 let's call it p1 of x is going to be the function evaluated at 1 plus the derivative of the function times x minus 1 so if we have that as our function well if at 1 would be just 1 over 1 which is 1 and if we have that as our function f f prime at will be just minus 1 over x squared so that f prime at 1 will be just minus 1 over 1 square which is 1 so it would be that so that the scalar polynomial is 1 plus minus 1 times x minus 1 and distributing these 2 this 2 the 0 minus x minus and minus that becomes a plus plus one so it is going to be 2 minus x so this is the taylor polynomial at a equals 1 at that point that is important that a equals 1 not now this second the quadratic approximation or the taylor polynomial of degree 2 this is the taylor poly of degree 2 is going to be be of this polynomial of degree one the taylor polynomial of degree two is going to be the taylor polynomial of degree one or if we are keeping the same point at a equals 1 plus the derivative of the second derivative the additional terms is the second derivative at 1 divided by 2 times 6 minus 1 squared so we have this as i have prime f double prime will be equal to well you will get a this is a this is the same as minus x to the minus two so that it will be minus minus two times x to the minus three and this is uh two times x minus three so that all these is value uh that evaluated at the second derivative evaluated at one would be just 2.

03:43 2 times 1 to the minus 3 which is 1 so it would be just 2.

03:49 So that this second degree polynomial would be the first degree polynomial plus 2 divided by 2 times x minus 1 squared...

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