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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Question
![**Problem Statement:**
**Find the equation of the tangent line to the graph of \( f(x) = \frac{2}{x} \) at \(\left( 4, \frac{1}{2} \right) \).**
---
**Solution:**
The slope of the tangent line to the graph of \( f(x) \) at a point is given by the derivative of \( f(x) \) evaluated at that point.
First, let's find the derivative of \( f(x) = \frac{2}{x} \).
Rewrite \( f(x) \) as \( f(x) = 2x^{-1} \).
Using the power rule, \( f'(x) = -2x^{-2} = -\frac{2}{x^2} \).
Next, we evaluate \( f'(x) \) at \( x = 4 \):
\[ f'(4) = -\frac{2}{4^2} = -\frac{2}{16} = -\frac{1}{8} \]
So, the slope of the tangent line at \( \left( 4, \frac{1}{2} \right) \) is \( -\frac{1}{8} \).
Now, using the point-slope form of a linear equation \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point of tangency:
\[ y - \frac{1}{2} = -\frac{1}{8}(x - 4) \]
Simplify this equation to get the tangent line's equation:
\[ y - \frac{1}{2} = -\frac{1}{8} x + \frac{1}{2} \]
Adding \( \frac{1}{2} \) to both sides results in:
\[ y = -\frac{1}{8} x + 1 \]
Therefore, the equation of the tangent line is:
\[ \boxed{y = -\frac{1}{8} x + 1} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F708b5043-edcc-41eb-8461-039a390d64a5%2Fc77f48e6-301b-45c1-82b0-8f4e64cf599c%2Fmj8qwi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
**Find the equation of the tangent line to the graph of \( f(x) = \frac{2}{x} \) at \(\left( 4, \frac{1}{2} \right) \).**
---
**Solution:**
The slope of the tangent line to the graph of \( f(x) \) at a point is given by the derivative of \( f(x) \) evaluated at that point.
First, let's find the derivative of \( f(x) = \frac{2}{x} \).
Rewrite \( f(x) \) as \( f(x) = 2x^{-1} \).
Using the power rule, \( f'(x) = -2x^{-2} = -\frac{2}{x^2} \).
Next, we evaluate \( f'(x) \) at \( x = 4 \):
\[ f'(4) = -\frac{2}{4^2} = -\frac{2}{16} = -\frac{1}{8} \]
So, the slope of the tangent line at \( \left( 4, \frac{1}{2} \right) \) is \( -\frac{1}{8} \).
Now, using the point-slope form of a linear equation \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point of tangency:
\[ y - \frac{1}{2} = -\frac{1}{8}(x - 4) \]
Simplify this equation to get the tangent line's equation:
\[ y - \frac{1}{2} = -\frac{1}{8} x + \frac{1}{2} \]
Adding \( \frac{1}{2} \) to both sides results in:
\[ y = -\frac{1}{8} x + 1 \]
Therefore, the equation of the tangent line is:
\[ \boxed{y = -\frac{1}{8} x + 1} \]
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