/*! 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 73 Determine the point(s) at which ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 73

Determine the point(s) at which the graph of the function has a horizontal tangent line. \(f(x)=\frac{x^{2}}{x-1}\)

Short Answer

Expert verified
The function \(f(x) = \frac{x^{2}}{x-1}\) has horizontal tangents at \(x = 0\) and \(x = 2\).

Step by step solution

01

Compute the Derivative

First need to compute the derivative of the given function, using the quotient rule which is \((\frac{u}{v})'=\frac{u'v-uv'}{v^{2}}\). Here \(u = x^2\) and \(v = x - 1\), so compute \(u'\) and \(v'\) first. \(u'= 2x\) and \(v' = 1\). Now we can compute the derivative: \(f'(x)=\frac{(2x)(x - 1) - (x^2)(1)}{(x - 1)^2} = \frac{2x^2 - 2x - x^2}{(x - 1)^2} = \frac{x^2 - 2x}{(x - 1)^2}\).
02

Find Where the Derivative Equals 0

Now to find the x-values for which the derivative is 0, meaning \(\frac{x^2 - 2x}{(x - 1)^2} = 0\). The numerator of a fraction must equal 0 for the entire fraction to be 0, so this simplifies to finding the roots of the equation \(x^2 - 2x = 0\).
03

Solve the Quadratic Equation

Now need to solve \(x^2 - 2x = 0\). This is a simple quadratic equation, which can be factored to \(x(x - 2) = 0\). Setting each factor equal to 0 gives the possible solutions \(x = 0\) and \(x = 2\).
04

Check the Validity of the Solutions

The potential solutions \(x = 0\) and \(x = 2\) must be checked against the domain of the original f(x) function. However, the function is undefined at \(x = 1\), but is defined everywhere else including our solutions.
05

Final Answer

The function \(f(x) = \frac{x^{2}}{x-1}\) has horizontal tangent lines at \(x = 0\) and \(x = 2\).

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.

Understanding Derivatives
Derivatives are a fundamental concept in calculus. They measure how a function changes as its input changes. Think of it as the rate of change or the slope of a function at a given point.
For a function \( f(x) \), the derivative \( f'(x) \) gives the slope of the tangent line at any point \( x \). This is crucial for understanding how the function behaves locally.
  • To compute a derivative, you need to apply specific rules, depending on the function's form.
  • Common rules include the power rule, product rule, quotient rule, and chain rule.
In this exercise, we'll apply the quotient rule to find the derivative of a rational function.
Finding Horizontal Tangent Lines
A horizontal tangent line occurs at points where the derivative of a function is zero. This means there's no slope at that point鈥攖he graph is flat like a horizontal line.
Why is this important? Horizontal tangents indicate peaks, troughs, or level sections of the graph, helping us understand the behavior of the function:
  • Solving \( f'(x) = 0 \) will give the x-values where the graph has horizontal tangents.
  • These x-values often show where a function changes direction.
In our problem, we find these points by setting the derived expression to zero, indicating potential maxima or minima.
Applying the Quotient Rule
The quotient rule is used to find the derivative of a quotient or division of two functions. If you have a function \( f(x) = \frac{u(x)}{v(x)} \), the quotient rule states that:\[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]Here's how you can break it down:
  • Identify the top and bottom functions: \( u(x) \) and \( v(x) \).
  • Find their derivatives: \( u'(x) \) and \( v'(x) \).
  • Plug these into the formula.
For the function \( f(x) = \frac{x^2}{x-1} \), applying the quotient rule helps us find the derivative needed to solve the problem.
Solving Quadratic Equations
Quadratic equations appear frequently in mathematics and are typically in the form \( ax^2 + bx + c = 0 \). The solutions are the x-values where the equation is true.
To solve these, you can:
  • Factor the equation if possible, by finding numbers that multiply to \( ac \) and add to \( b \).
  • Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
  • Complete the square, if necessary.
In our exercise, the equation \( x^2 - 2x = 0 \) is factored easily to \( x(x - 2) = 0 \), giving solutions \( x = 0 \) and \( x = 2 \), indicating the points of horizontal tangency.

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

The displacement from equilibrium of an object in harmonic motion on the end of a spring is \(y=\frac{1}{3} \cos 12 t-\frac{1}{4} \sin 12 t\) where \(y\) is measured in feet and \(t\) is the time in seconds. Determine the position and velocity of the object when \(t=\pi / 8 .\)

A ladder 25 feet long is leaning against the wall of a house (see figure). The base of the ladder is pulled away from the wall at a rate of 2 feet per second. (a) How fast is the top of the ladder moving down the wall when its base is 7 feet, 15 feet, and 24 feet from the wall? (b) Consider the triangle formed by the side of the house, the ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall. (c) Find the rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 7 feet from the wall.

Find the slope of the tangent line to the graph at the given point. Cissoid: \((4-x) y^{2}=x^{3}\) Point: \((2,2)\)

A baseball diamond has the shape of a square with sides 90 feet long (see figure). A player running from second base to third base at a speed of 28 feet per second is 30 feet from third base. At what rate is the player's distance \(s\) from home plate changing?

Find the second derivative of the function. $$ f(x)=\frac{1}{x-2} $$
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Decision Maths

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.