Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![## Problem Statement
Determine the equation of the line tangent to the curve \( y = 3 \left( \frac{x + 4}{x - 4} \right)^2 \) at the point \((6, 75)\).
## Solution
To find the equation of the tangent to the curve at a given point, we need:
1. The slope of the curve at that point.
2. The coordinates of the point of tangency.
### Step 1: Find the Derivative
First, we'll find the derivative of the given function to determine the slope of the tangent line. The function is:
\[ y = 3 \left( \frac{x + 4}{x - 4} \right)^2 \]
Let \( u = \frac{x + 4}{x - 4} \), then \( y = 3u^2 \).
Using the chain rule:
\[ \frac{dy}{dx} = 3 \cdot 2u \cdot \frac{du}{dx} = 6u \cdot \frac{du}{dx} \]
Next, find \( \frac{du}{dx} \):
\[ u = \frac{x + 4}{x - 4} \]
Using the quotient rule:
\[ \frac{du}{dx} = \frac{(x - 4) \cdot 1 - (x + 4) \cdot 1}{(x - 4)^2} = \frac{x - 4 - x - 4}{(x - 4)^2} = \frac{-8}{(x - 4)^2} \]
So,
\[ \frac{dy}{dx} = 6u \cdot \frac{-8}{(x - 4)^2} = \frac{-48 \left( \frac{x + 4}{x - 4} \right)}{(x - 4)^2} = \frac{-48(x + 4)}{(x - 4)^3} \]
### Step 2: Evaluate the Slope at the Given Point
Substitute \( x = 6 \) into the expression for the derivative:
\[ u = \frac{6 + 4}{6 - 4} = \frac{10}{2} = 5 \]
Therefore,
\[ \frac{dy}{dx} \Big](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F09277b3e-bfbb-4704-b58b-5d83ab5b8c9a%2F3cc317d0-3d44-4660-80f9-ba65565d4686%2Fe3st29k_processed.png&w=3840&q=75)
Transcribed Image Text:## Problem Statement
Determine the equation of the line tangent to the curve \( y = 3 \left( \frac{x + 4}{x - 4} \right)^2 \) at the point \((6, 75)\).
## Solution
To find the equation of the tangent to the curve at a given point, we need:
1. The slope of the curve at that point.
2. The coordinates of the point of tangency.
### Step 1: Find the Derivative
First, we'll find the derivative of the given function to determine the slope of the tangent line. The function is:
\[ y = 3 \left( \frac{x + 4}{x - 4} \right)^2 \]
Let \( u = \frac{x + 4}{x - 4} \), then \( y = 3u^2 \).
Using the chain rule:
\[ \frac{dy}{dx} = 3 \cdot 2u \cdot \frac{du}{dx} = 6u \cdot \frac{du}{dx} \]
Next, find \( \frac{du}{dx} \):
\[ u = \frac{x + 4}{x - 4} \]
Using the quotient rule:
\[ \frac{du}{dx} = \frac{(x - 4) \cdot 1 - (x + 4) \cdot 1}{(x - 4)^2} = \frac{x - 4 - x - 4}{(x - 4)^2} = \frac{-8}{(x - 4)^2} \]
So,
\[ \frac{dy}{dx} = 6u \cdot \frac{-8}{(x - 4)^2} = \frac{-48 \left( \frac{x + 4}{x - 4} \right)}{(x - 4)^2} = \frac{-48(x + 4)}{(x - 4)^3} \]
### Step 2: Evaluate the Slope at the Given Point
Substitute \( x = 6 \) into the expression for the derivative:
\[ u = \frac{6 + 4}{6 - 4} = \frac{10}{2} = 5 \]
Therefore,
\[ \frac{dy}{dx} \Big
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