/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 Quadratic approximations $$ ... [FREE SOLUTION] | 91Ó°ÊÓ

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Quadratic approximations $$ \begin{array}{l}{\text { a. Let } Q(x)=b_{0}+b_{1}(x-a)+b_{2}(x-a)^{2} \text { be a quadratic }} \\ {\text { approximation to } f(x) \text { at } x=a \text { with the properties: }}\end{array} $$ $$ \begin{aligned} \text { i) } Q(a) &=f(a) \\ \text { ii) } Q^{\prime}(a) &=f^{\prime}(a) \\ \text { iii) } Q^{\prime \prime}(a) &=f^{\prime \prime}(a) \end{aligned} $$ $$ \text{Determine the coefficients}b_{0}, b_{1}, \text { and } b_{2} $$ $$ \begin{array}{l}{\text { b. Find the quadratic approximation to } f(x)=1 /(1-x) \text { at }} \\ {\quad x=0 .} \\ {\text { c. } \operatorname{Graph} f(x)=1 /(1-x) \text { and its quadratic approximation at }} \\ {x=0 . \text { Then zoom in on the two graphs at the point }(0,1) \text { . }} \\ {\text { Comment on what you see. }}\end{array} $$ $$ \begin{array}{l}{\text { d. Find the quadratic approximation to } g(x)=1 / x \text { at } x=1} \\ {\text { Graph } g \text { and its quadratic approximation together. Comment }} \\ {\text { on what you see. }}\end{array} $$ $$ \begin{array}{l}{\text { e. Find the quadratic approximation to } h(x)=\sqrt{1+x} \text { at }} \\ {x=0 . \text { Graph } h \text { and its quadratic approximation together. }} \\ {\text { Comment on what you see. }} \\\ {\text { f. What are the linearizations of } f, g, \text { and } h \text { at the respective }} \\ {\text { points in parts (b), (d), and (e)? }}\end{array} $$

Short Answer

Expert verified
Compute the quadratic approximation coefficients: - \( b_0 = f(a) \) - \( b_1 = f'(a) \) - \( b_2 = \frac{f''(a)}{2} \).

Step by step solution

01

Determine coefficients of Q(x)

Given the quadratic approximation form \( Q(x) = b_0 + b_1(x-a) + b_2(x-a)^2 \) and the conditions:1. \( Q(a) = f(a) \): Substituting \( x = a \) gives \( b_0 = f(a) \).2. \( Q'(a) = f'(a) \): Differentiating \( Q(x) \) we get \( Q'(x) = b_1 + 2b_2(x-a) \), substituting \( x = a \) gives \( b_1 = f'(a) \).3. \( Q''(a) = f''(a) \): Differentiating again \( Q'(x) \) gives \( Q''(x) = 2b_2 \), substituting \( x = a \) gives \( 2b_2 = f''(a) \) so \( b_2 = \frac{f''(a)}{2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Taylor Series
The Taylor Series is a powerful tool for approximating functions using polynomials. It's especially useful when you need to make complex functions simpler near a specific point. Essentially, a Taylor Series expands a function into an infinite sum of terms, which are derived from the function's derivatives at a single point. This approach relies heavily on the idea of function continuity and differentiability. By finding derivatives up to a certain order and evaluating them at a point, you create a polynomial that approximates the function around that area.
  • The quadratic approximation is a special case of the Taylor Series, using only up to the second derivative of the function.
  • It helps understand how the function behaves near a particular point by considering its slope and curvature.
  • This approximation is expressed as: \[Q(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2\]
So, when you see a function like a Taylor Series expansion, think of it as peeling back layers of a function to know its behavior near a point.
Derivatives
Understanding derivatives is crucial for grasping quadratic approximations and Taylor Series. A derivative represents the rate at which a function is changing at any given point, effectively acting as a function's instant slope.For quadratic approximations, the first and second derivatives play a key role:
  • The first derivative, \(f'(x)\), informs us about the linear behavior of the function, like its slope.
  • The second derivative, \(f''(x)\), tells us about the function's curvature, i.e., how the slope itself is changing.
By ensuring the approximation meets the function at the same value, slope, and curvature at a point, we create a polynomial that closely mirrors the function's behavior near that point. A savvy grasp of derivatives empowers you to dissect and predict function behavior effectively.
Function Graphing
Graphing a function and its approximation can reveal a lot about their relationship. By visualizing these functions, you can see how well an approximation fits the original function at the point of interest. Let's consider some helpful tips when comparing graphs:
  • Graph the original function first to understand its basic shape and behavior.
  • Then, plot the quadratic approximation alongside the original. Notice how close the two are within a small neighborhood around the approximation point.
  • As you "zoom in" on the point of approximation, observe that the two graphs should almost overlap if the approximation is precise.
This observation aligns with the underlying mathematics, where achieving identical values, slopes, and curvatures at the point of interest ensures a good fit for the quadratic approximation. Regular practice of graphing these relationships strengthens your understanding of function behaviors and approximation accuracy.

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Most popular questions from this chapter

A building's shadow On a morning of a day when the sun will pass directly overhead, the shadow of an 80-ft building on level ground is 60 \(\mathrm{ft}\) long. At the moment in question, the angle \(\theta\) the sun makes with the ground is increasing at the rate of \(0.27^{\circ} / \mathrm{min}\) . At what rate is the shadow decreasing? (Remember to use radians. Express your answer in inches per minute, to the nearest tenth.)

Boring a cylinder The mechanics at Lincoln Automotive are reboring a 6 -in.-deep cylinder to fit a new piston. The machine they are using increases the cylinder's radius one-thousandth of an inch every 3 min. How rapidly is the cylinder volume increasing when the bore (diameter) is 3.800 in.?

Temperature and the period of a pendulum For oscillations of small amplitude (short swings), we may safely model the relationship between the period \(T\) and the length \(L\) of a simple pendulum with the equation $$T=2 \pi \sqrt{\frac{L}{g}}$$ where \(g\) is the constant acceleration of gravity at the pendulum's location. If we measure \(g\) in centimeters per second squared, we measure \(L\) in centimeters and \(T\) in seconds. If the pendulum is made of metal, its length will vary with temperature, either increasing or decreasing at a rate that is roughly proportional to \(L .\) In symbols, with \(u\) being temperature and \(k\) the proportionality constant, $$\frac{d L}{d u}=k L$$ Assuming this to be the case, show that the rate at which the period changes with respect to temperature is \(k T / 2 .\)

Use a CAS to perform the following steps for the functions \begin{equation} \begin{array}{l}{\text { a. Plot } y=f(x) \text { to see that function's global behavior. }} \\ {\text { b. Define the difference quotient } q \text { at a general point } x, \text { with }} \\ {\text { general step size } h .} \\\ {\text { c. Take the limit as } h \rightarrow 0 . \text { What formula does this give? }} \\ {\text { d. Substitute the value } x=x_{0} \text { and plot the function } y=f(x)} \\ \quad {\text { together with its tangent line at that point. }} \\ {\text { e. Substitute various values for } x \text { larger and smaller than } x_{0} \text { into }} \\ \quad {\text { the formula obtained in part (c). Do the numbers make sense }} \\ \quad {\text { with your picture? }} \\ {\text { f. Graph the formula obtained in part (c). What does it mean }} \\ \quad {\text { when its values are negative? Zero? Positive? Does this make }} \\ \quad {\text { sense with your plot from part (a)? Give reasons for your }} \\ \quad {\text { answer. }}\end{array} \end{equation} $$f(x)=x^{3}+x^{2}-x, x_{0}=1$$

If \(x^{2} y^{3}=4 / 27\) and \(d y / d t=1 / 2,\) then what is \(d x / d t\) when \(x=2 ?\)

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