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Determine the equation of the line tangent to the curve y = 3| : 3(x+4)² y = at the point (6,75).

Determine the equation of the line tangent to the curve y = 3| : 3(x+4)² y = at the point (6,75).

Calculus: Early Transcendentals
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
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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