/*! 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 64 Find an equation of the line \(l... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 64

Find an equation of the line \(l\) tangent to the graph of \(f\) at the given point. \(f(x)=\left(1+x^{1 / 3}\right)^{2 / 3} ;(-8,1)\)

Short Answer

Expert verified
The equation of the tangent line is \( y = -\frac{1}{18}x + \frac{5}{9} \).

Step by step solution

01

Find the Derivative

To find the equation of the tangent line, we first need the derivative of the function \( f(x) = (1 + x^{1/3})^{2/3} \) to determine the slope. Let \( u = 1 + x^{1/3} \). Then, \( f(x) = u^{2/3} \). The derivative \( f'(x) \) is found using the chain and power rules.First, differentiate \( u^{2/3} \):\[ \frac{d}{du}(u^{2/3}) = \frac{2}{3}u^{-1/3} \]Then, differentiate \( u = 1 + x^{1/3} \):\[ \frac{d}{dx}(1 + x^{1/3}) = \frac{1}{3}x^{-2/3} \]Now, apply the chain rule:\[ f'(x) = \frac{2}{3}(1 + x^{1/3})^{-1/3} \cdot \frac{1}{3}x^{-2/3} = \frac{2}{9} \cdot (1 + x^{1/3})^{-1/3} \cdot x^{-2/3} \].
02

Evaluate the Derivative at the Given Point

To find the slope of the tangent line at the point \((-8, 1)\), evaluate \( f'(x) \) at \( x = -8 \).Calculate: \( 1 + (-8)^{1/3} = 1 - 2 = -1 \), then \[ f'(-8) = \frac{2}{9} \cdot (-1)^{-1/3} \cdot (-8)^{-2/3}. \]Since \((-1)^{-1/3} = -1 \) and \((-8)^{-2/3} = \frac{1}{4} \), then:\[ f'(-8) = \frac{2}{9} \cdot (-1) \cdot \frac{1}{4} = -\frac{1}{18}. \]The slope of the tangent line is \(-\frac{1}{18}\).
03

Use the Point-Slope Form to Write the Equation

With the point \((-8, 1)\) and the slope \( m = -\frac{1}{18} \), use the point-slope form of a line:\[ y - y_1 = m(x - x_1) \]Substitute \( (x_1, y_1) = (-8, 1) \) and \( m = -\frac{1}{18} \):\[ y - 1 = -\frac{1}{18}(x + 8). \]Simplify:\[ y - 1 = -\frac{1}{18}x - \frac{8}{18} \]\[ y = -\frac{1}{18}x - \frac{4}{9} + 1 \]\[ y = -\frac{1}{18}x + \frac{5}{9}. \]This is the equation of the tangent line.

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 91影视!

Key Concepts

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

Tangent Line
A tangent line is a straight line that touches a curve at a single point. This line represents the instantaneous direction in which the curve is heading at that specific contact point. Think of it like a snapshot of the curve at just one moment. The key features of a tangent line are:
  • It only intersects the curve at one specific point.
  • At the point of contact, it has the same slope as the curve.
  • The tangent line does not cross through the curve nearby the contact point but merely "touches" it.
To find the equation of a tangent line, we must calculate its slope, which is the same as the slope of the curve at the contact point. This involves using calculus and particularly derivatives, to determine how the function behaves at that point. Once we have the slope, we can use a point-slope form of a line to write the complete equation of the tangent.
Derivative Calculation
When discussing tangent lines, the derivative of a function becomes very important. A derivative is a tool in calculus that gives us the rate at which a function is changing at any given point. In simple terms, it tells us the slope of the tangent line to the curve at a specific point.

To find the derivative of a function, numerous techniques can be used. In the case of the function given in the exercise, we used the chain and power rules to differentiate. This involves:
  • Identifying components of the function that can be simplified.
  • Applying mathematical rules like the power rule, which states that the derivative of \(x^n\) is \(n \cdot x^{n-1}\).
  • Using the chain rule to differentiate composite functions, which are functions within functions.
Having calculated the derivative, we plug the specific x-value from the given point into the derivative to find the slope of the tangent line at that exact point.
Chain Rule
The chain rule is a fundamental tool in calculus used for finding the derivative of composite functions. A composite function is essentially a function inside another function, such as our example with \(f(x) = (u^{2/3})\) where \(u = 1 + x^{1/3}\).To apply the chain rule, you must:
  • Identify the inner and outer functions. In this case, the outer function is \(u^{2/3}\), and the inner function is \(1 + x^{1/3}\).
  • Differentiating each function separately.

  • Multiply the derivative of the outer function by the derivative of the inner function. In our example, applying the chain rule means we first differentiate \(u^{2/3}\) and then multiply it by the derivative of \(1 + x^{1/3}\).
The result is a new function, which is the derivative of the composite function. This derivative function will help us determine the slope of the tangent line at the desired point.
Point-Slope Form of a Line
The point-slope form is a way to write the equation of a line when you know the slope and a point on the line. It is particularly useful when you have already calculated the slope using the derivative of a function.The point-slope form is given by the equation:\[ y - y_1 = m(x - x_1) \]where:
  • \(m\) is the slope of the line.
  • \((x_1, y_1)\) is the point on the line.
In the context of our exercise, we had a slope \(m\) of \(-\frac{1}{18}\) and the point \((-8, 1)\). By plugging these into the point-slope form, we derived the equation of the tangent line.This method is straightforward and quick once the slope is known, making it an efficient tool for finding the equation of tangent lines in calculus.

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

Find the velocity and acceleration of an object moving along the \(x\) axis and having the given position function. $$ f(t)=3-1 / t^{2} $$

The outside surface of a hemispherical dome of radius 20 feet is to be given a coat of paint, \(\frac{1}{100}\) inch thick. Use differentials to approximate the volume of paint needed for the job. (Hint: Approximate the change in the volume of a hemisphere when the radius increases from 20 feet to \(20 \frac{1}{1200}\) feet.)

Find the derivative of the given function. $$ g(x)=-4 x^{-3}+2 \cos x $$

Find the velocity and acceleration of an object moving along the \(x\) axis and having the given position function. $$ f(t)=2 \sin t-3 \cos t $$

Use the results of this section to find the derivative of the given function at the given numbers. $$ f(t)=\cos t ; a=0,-\pi / 3 $$
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation
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.