/*! 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 4 In Exercises \(1-4,\) find the l... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises \(1-4,\) find the linearization \(L(x)\) of \(f(x)\) at \(x=a\) $$ f(x)=\sqrt[3]{x}, \quad a=-8 $$

Short Answer

Expert verified
The linearization \( L(x) \) of \( f(x)=\sqrt[3]{x} \) at \( x=-8 \) is \( L(x) = \frac{1}{12}x - \frac{4}{3} \).

Step by step solution

01

Understand Linearization

Linearization is the process of approximating a function near a given point using a tangent line. The formula for linearization at a point \( x = a \) is given by \[ L(x) = f(a) + f'(a)(x-a) \], where \( f(a) \) is the value of the function at \( x = a \) and \( f'(a) \) is the derivative of the function evaluated at \( a \).
02

Calculate \( f(a) \)

Find the value of the function \( f(x) = \sqrt[3]{x} \) at \( x = -8 \).\[ f(-8) = \sqrt[3]{-8} = -2 \].
03

Find the Derivative \( f'(x) \)

Differentiate the function \( f(x) = \sqrt[3]{x} \) using the power rule.\[ f(x) = x^{1/3} \rightarrow f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}} \].
04

Evaluate \( f'(a) \)

Substitute \( x = -8 \) into \( f'(x) = \frac{1}{3\sqrt[3]{x^2}} \) to find \( f'(-8) \).\[ f'(-8) = \frac{1}{3\sqrt[3]{(-8)^2}} = \frac{1}{3\sqrt[3]{64}} = \frac{1}{12} \].
05

Formulate the Linearization \( L(x) \)

Use the linearization formula \( L(x) = f(a) + f'(a)(x-a) \) to find \( L(x) \) at \( a = -8 \).\[ L(x) = -2 + \frac{1}{12}(x + 8) \].
06

Simplify the Linearization

Simplify the expression for \( L(x) \).\[ L(x) = -2 + \frac{1}{12}x + \frac{8}{12} = -2 + \frac{1}{12}x + \frac{2}{3} \].\[ L(x) = \frac{1}{12}x - \frac{4}{3} \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus, focusing on how functions change, or their rate of change. It is essentially about finding the derivative of a function. This tells us how the function behaves at any given point. The derivative can be thought of as the slope of the tangent line to the function at a particular point. For a function like \( f(x) = \sqrt[3]{x} \), differentiation helps us understand how steep the curve is at different points. By differentiating this function, we know precisely how quickly it grows or decreases as \( x \) changes, which is crucial for understanding its behavior.
Tangent Line
A tangent line is a straight line that touches a curve at one point, without crossing it. At this point, the line has the same slope as the curve. This means the tangent line is a great way to approximate the behavior of a curve very close to the touching point. For instance, if you have a curve representing \( f(x) = \sqrt[3]{x} \) and you look at the point where \( x = -8 \), the tangent line offers a snapshot, mimicking the function's behavior precisely at that point. Knowing the equation of this tangent line enables us to accurately estimate function values near our point of interest.
Power Rule
The Power Rule is a quick technique for finding the derivative of a function of the form \( x^n \). According to the Power Rule, the derivative of \( x^n \) is \( nx^{n-1} \). This is particularly useful because it simplifies the differentiation process, allowing us to efficiently find how a function changes. Let's apply this rule to our example, \( f(x) = x^{1/3} \). By applying the Power Rule, we find the derivative: \( f'(x) = \frac{1}{3}x^{-2/3} \). Here, the exponent "1/3" becomes the coefficient, and we subtract one from the exponent to yield "-2/3". This calculation helps determine the slope of the tangent line, which is vital for linear approximation.
Function Approximation
Function approximation is a technique used to find simple expressions that can closely predict or estimate values of a more complex function. Linearization is one of the methods used for this purpose, employing the equation of the tangent line as the approximation tool.By computing \( L(x) = f(a) + f'(a)(x-a) \), we take the exact function, like \( \sqrt[3]{x} \), and replace it with a linear function that gives similar results near the point of interest, \( x = -8 \). This form uses both the function value \( f(a) \) and the derivative \( f'(a) \), simplifying complex calculations into easier linear ones, accurately reflecting the local behavior of the function. Linear approximations are especially beneficial when evaluating functions that are difficult to compute directly.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use a CAS to perform the following steps in Exercises \(77-84\) . a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P .\) c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P\) . Then plot the implicit curve and tangent line together on a single graph. $$ 2 y^{2}+(x y)^{1 / 3}=x^{2}+2, \quad P(1,1) $$

In Exercises \(47-56,\) verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point. $$ y=2 \sin (\pi x-y), \quad(1,0) $$

In Exercises \(31-36,\) each function \(f(x)\) changes value when \(x\) changes from \(x_{0}\) to \(x_{0}+d x .\) Find a. the change \(\Delta f=f\left(x_{0}+d x\right)-f\left(x_{0}\right)\) b. the value of the estimate \(d f=f^{\prime}\left(x_{0}\right) d x ;\) and c. the approximation error \(|\Delta f-d f|\) $$ f(x)=x^{3}-2 x+3, \quad x_{0}=2, \quad d x=0.1 $$

In Exercises \(67-70\) , use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\) . Perform the following steps: a. Plot the function \(f\) over \(I .\) b. Find the linearization \(L\) of the function at the point \(a\) . c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta > 0\) as you can, satisfying $$ |x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon $$ for \(\epsilon=0.5,0.1,\) and 0.01 . Then check graphically to see if your \(\delta\) -estimate holds true. $$ f(x)=\frac{x-1}{4 x^{2}+1}, \quad\left[-\frac{3}{4}, 1\right], \quad a=\frac{1}{2} $$

In Exercises \(37-42,\) write a differential formula that estimates the given change in volume or surface area. The change in the volume \(V=x^{3}\) of a cube when the edge lengths change from \(x_{0}\) to \(x_{0}+d x\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.