/*! 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 13 Show that the linearization of \... [FREE SOLUTION] | 91Ó°ÊÓ

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Show that the linearization of \(f(x)=(1+x)^{k}\) at \(x=0\) is \(L(x)=1+k x\).

Short Answer

Expert verified
The linearization of \(f(x)=(1+x)^{k}\) at \(x=0\) is \(L(x)=1+kx\).

Step by step solution

01

Understand Linearization

Linearization is the process of approximating a function using a linear function (a straight line) near a specific point. For a function \( f(x) \), its linearization at a point \( a \) is given by the formula \( L(x) = f(a) + f'(a)(x-a) \). In this problem, we will linearize the function \( f(x) = (1+x)^k \) at the point \( x = 0 \).
02

Compute f(a)

First, determine \( f(0) \) for the given function \( f(x) = (1+x)^{k} \). Substituting \( x = 0 \) into the function, we get: \[ f(0) = (1 + 0)^k = 1^k = 1. \]
03

Find the derivative f'(x)

Calculate the derivative \( f'(x) \) of the function \( f(x) = (1+x)^{k} \). Using the power rule, the derivative is \[ f'(x) = k(1+x)^{k-1}. \]
04

Compute f'(a)

Evaluate the derivative at the point \( x = 0 \): \[ f'(0) = k(1+0)^{k-1} = k(1) = k. \]
05

Formulate the Linearization L(x)

Substitute \( f(0) \) and \( f'(0) \) into the linearization formula \( L(x) = f(0) + f'(0)(x-0) \): \[ L(x) = 1 + kx. \]
06

Verify the Result

Check if any computation steps were missed or if there's a mistake in the logic. The derived linearization \( L(x) = 1 + kx \) matches the given target, confirming the solution is correct.

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Key Concepts

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

Differential Calculus
Differential calculus is a branch of mathematics focused on how things change. More specifically, it deals with the concept of the derivative, which tells us the rate at which a quantity changes. In the context of the linearization exercise, we are looking at how the function \(f(x) = (1+x)^k\) changes near a specific point, \(x = 0\).

The derivative gives us the slope of the tangent line at any point on a function. This is essential because the slope of this tangent line approximates the behavior of the function at that point. Through differential calculus, we can simplify complex problems by focusing on small, instantaneous changes, helping us understand the behavior of functions on an infinitesimal level.
Power Rule
The power rule is a fundamental principle in differential calculus used to find derivatives of functions of the form \(x^n\), where \(n\) is any real number. The rule states that if \(f(x) = x^n\), then its derivative \(f'(x)\) is \(nx^{n-1}\). In our exercise, the function \(f(x) = (1+x)^k\) requires using the power rule to calculate its derivative.

When we applied the power rule to find the derivative of \((1+x)^k\), we treated \(1+x\) as a single variable raised to the exponent \(k\). Thus, the derivative becomes \(k(1+x)^{k-1}\). This straightforward application of the power rule makes it an incredibly powerful tool for finding the rate of change of polynomial functions.
Function Approximation
Function approximation is a technique used to simplify complex functions by estimating them with simpler ones, like a linear function. In this exercise, we are approximating the function \(f(x) = (1+x)^k\) with a straight line, \(L(x) = 1 + kx\), near \(x = 0\).

By approximating a complex function with a simpler one, such as a line, we can easily analyze its behavior near a specific point. This is especially helpful in solving real-world problems where precision isn't as crucial, or when computational resources are limited. Function approximation, therefore, allows us to break down otherwise complex problems into simpler, more manageable parts.
Derivatives
Derivatives are a cornerstone concept in differential calculus, representing how a function's output value changes as its input changes. They are closely associated with the concept of linearization, as seen in our exercise.

For the function \(f(x) = (1+x)^k\), its derivative \(f'(x) = k(1+x)^{k-1}\) tells us about the slope of the curve at any given point. In linearization, this derivative becomes crucial as it defines the slope of the linear approximation \(L(x)\). By substituting the derivative into the linearization equation, we attain a simple line that approximates \(f(x)\) near \(x=0\), showing the practical utility of derivatives in simplifying and understanding functions.

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Most popular questions from this chapter

Find \(d y / d t\) $$y=(t \tan t)^{10}$$

What happens if you can write a function as a composite in different ways? Do you get the same derivative each time? The Chain Rule says you should. Try it with the functions. Find \(d y / d x\) if \(y=x^{3 / 2}\) by using the Chain Rule with \(y\) as a composite of a. \(y=u^{3}\) and \(u=\sqrt{x}\) b. \(y=\sqrt{u}\) and \(u=x^{3}\)

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 \(k T / 2\)

Find both \(d y / d x\) (treating \(y\) as a differentiable function of \(x\) ) and \(d x / d y\) (treating \(x\) as a differentiable function of \(y\) ). How do \(d y / d x\) and \(d x / d y\) seem to be related? Explain the relationship geometrically in terms of the graphs. $$x y^{3}+x^{2} y=6$$

Find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=u^{5}+1, \quad u=g(x)=\sqrt{x}, \quad x=1$$

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