/*! 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 Find the slope of the function's... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 13

Find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. \begin{equation} g(x)=\frac{x}{x-2}, \quad(3,3) \end{equation}

Short Answer

Expert verified
The slope is -2 and the tangent line is \( y = -2x + 9 \).

Step by step solution

01

Differentiate the Function

We need to find the derivative of the function \( g(x) = \frac{x}{x-2} \) to determine the slope of the tangent line. Use the quotient rule to differentiate. The quotient rule states: If \( u(x) = \frac{f(x)}{h(x)} \), then the derivative \( u'(x) = \frac{f'(x)h(x) - f(x)h'(x)}{(h(x))^2} \).For our function, let \( f(x) = x \) and \( h(x) = x - 2 \), so \( f'(x) = 1 \) and \( h'(x) = 1 \). Applying the quotient rule gives:\[g'(x) = \frac{(1)(x-2) - x(1)}{(x-2)^2} = \frac{x-2-x}{(x-2)^2} = \frac{-2}{(x-2)^2}\]
02

Evaluate the Derivative at the Given Point

Next, evaluate the derivative at \( x = 3 \) to find the slope of the tangent line at the point \( (3, 3) \). Substitute \( x = 3 \) in \( g'(x) = \frac{-2}{(x-2)^2} \):\[g'(3) = \frac{-2}{(3-2)^2} = \frac{-2}{1} = -2\]
03

Write the Equation of the Tangent Line

With the slope \( m = -2 \) and the point \( (3, 3) \), use the point-slope form to write the equation of the tangent line. The point-slope form is \( y - y_1 = m(x - x_1) \).Substitute \( m = -2 \), \( x_1 = 3 \), and \( y_1 = 3 \):\[y - 3 = -2(x - 3)\]Simplify the equation:\[y = -2x + 6 + 3 = -2x + 9\]

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.

Quotient Rule
The quotient rule is a technique for finding the derivative of a function that is the ratio of two differentiable functions. It's particularly useful when dealing with functions of the form \( u(x) = \frac{f(x)}{h(x)} \). The formula for the quotient rule is:
  • \( u'(x) = \frac{f'(x)h(x) - f(x)h'(x)}{(h(x))^2} \)
To effectively use the quotient rule, identify the numerator \( f(x) \) and the denominator \( h(x) \) from your function. Differentiate both parts separately.
In our example, for \( g(x) = \frac{x}{x-2} \), the numerator is \( f(x) = x \) (thus \( f'(x) = 1 \)) and the denominator is \( h(x) = x - 2 \) (thus \( h'(x) = 1 \)).
Plug these into the quotient rule formula to get the derivative:
  • \( g'(x) = \frac{1 \cdot (x-2) - x \cdot 1}{(x-2)^2} \)
  • \( = \frac{x - 2 - x}{(x-2)^2} \)
  • \( = \frac{-2}{(x-2)^2} \)
This gives us the rate of change or the slope at any \( x \) point for the function \( g(x) \).
Derivative Evaluation
Once you've found the derivative of the function, the next step is to evaluate this derivative at a specific point. This step helps determine the slope of the tangent at the given point.
For our function \( g(x) = \frac{x}{x-2} \), we needed to evaluate the derivative at \( x = 3 \) to find the slope of the tangent line at the point \( (3, 3) \).
Plug \( x = 3 \) into the derivative formula \( g'(x) = \frac{-2}{(x-2)^2} \):
  • \( g'(3) = \frac{-2}{(3-2)^2} = \frac{-2}{1} = -2 \)
This calculation tells us that the slope of the tangent line at the point \( (3, 3) \) is \(-2\).
The derivative evaluation confirms how the function's rate of change behaves specifically at \( x = 3 \).
Point-Slope Form
After finding the slope of the tangent line, use the point-slope form to create the equation of the tangent line. This form is particularly useful when you have a specific point \((x_1, y_1)\) and a known slope \(m\).
  • The general formula is: \( y - y_1 = m(x - x_1) \)
Let's apply this to our exercise:
With a slope \( m = -2 \) and a known point \( (3, 3) \), substitute these into the formula.
  • \( y - 3 = -2(x - 3) \)
Simplify the expression by distributing and combining like terms to get the equation in the slope-intercept form \( y = mx + b \):
  • \( y = -2x + 6 + 3 \)
  • \( y = -2x + 9 \)
Thus, the equation of the tangent line at the point \( (3, 3) \) is \( y = -2x + 9 \).
This tells us the equation of the straight line that just touches the curve at that point, mirroring the curve's slope precisely there.

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

Clock's moving hands At what rate is the angle between a clock's minute and hour hands changing at 4 o'clock in the afternoon?

The effect of flight maneuvers on the heart The amount of work done by the heart's main pumping chamber, the left ventricle, is given by the equation $$ W=P V+\frac{V \delta v^{2}}{2 g} $$ where \(W\) is the work per unit time, \(P\) is the average blood pressure, \(V\) is the volume of blood pumped out during the unit of time, \(\delta(\) "delta") is the weight density of the blood, \(v\) is the average velocity of the exiting blood, and \(g\) is the acceleration of gravity. $$ \begin{array}{l}{\text { When } P, V, \delta, \text { and } v \text { remain constant, } W \text { becomes a function }} \\ {\text { of } g, \text { and the equation takes the simplified form }}\end{array} $$ $$ W=a+\frac{b}{g}(a, b \text { constant }) $$ As a member of NASA's medical team, you want to know how sensitive \(W\) is to apparent changes in \(g\) caused by flight maneuvers, and this depends on the initial value of \(g\) . As part of your investigation, you decide to compare the effect on \(W\) of a given change \(d g\) on the moon, where \(g=5.2 \mathrm{ft} / \mathrm{sec}^{2},\) with the effect the same change \(d g\) would have on Earth, where \(g=32 \mathrm{ft} / \mathrm{sec}^{2} .\) Use the simplified equation above to find the ratio of \(d W_{\mathrm{moon}}\) to \(d W_{\mathrm{Earth}}\)

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

Volume The radius \(r\) and height \(h\) of a right circular cone are related to the cone's volume \(V\) by the equation \(V=(1 / 3) \pi r^{2} h .\) a. How is \(d V / d t\) related to \(d h / d t\) if \(r\) is constant? b. How is \(d V / d t\) related to \(d r / d t\) if \(h\) is constant? c. How is \(d V / d t\) related to \(d r / d t\) and \(d h / d t\) if neither \(r\) nor \(h\) is constant?

In Exercises \(57-60,\) 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: $$ \begin{array}{l}{\text { a. Plot the function } f \text { over } I} \\ {\text { b. Find the linearization } L \text { of the function at the point } a \text { . }} \\ {\text { c. Plot } f \text { and } L \text { together on a single graph. }} \\ {\text { d. Plot the absolute error }|f(x)-L(x)| \text { over } I \text { and find its max- }} \\ {\text { imum value. }}\end{array} $$ $$ \begin{array}{l}{\text { e. From your graph in part (d), estimate as large a } \delta>0 \text { as you }} \\ {\text { can, satisfing }}\end{array} $$ $$ \begin{array}{c}{|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon} \\\ {\text { for } \epsilon=0.5,0.1, \text { and } 0.01 . \text { Then check graphically to see if }} \\ {\text { your } \delta \text { -estimate holds true. }}\end{array} $$ $$ f(x)=\frac{x-1}{4 x^{2}+1}, \quad\left[-\frac{3}{4}, 1\right], \quad a=\frac{1}{2} $$
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

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.