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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![## Determining an Equation for the Tangent Line
### Problem Statement:
Determine an equation for the tangent line to the graph at the given value:
\[ f(x) = 5^x, \, x = 2 \]
### Explanation:
To find the equation of the tangent line to the graph of the function \( f(x) = 5^x \) at the point where \( x = 2 \), follow these steps:
1. **Evaluate the function at \( x = 2 \)**:
\[ f(2) = 5^2 = 25 \]
2. **Find the derivative of the function**:
Using the chain rule for the derivative of an exponential function, \( f(x) = 5^x \), the derivative is:
\[ f'(x) = 5^x \ln(5) \]
3. **Evaluate the derivative at \( x = 2 \)**:
\[ f'(2) = 5^2 \ln(5) = 25 \ln(5) \]
4. **Write the equation of the tangent line**:
The equation of a line in point-slope form is given by:
\[ y - y_1 = m(x - x_1) \]
where \( m \) is the slope and \( (x_1, y_1) \) is the point of tangency.
In this case, \( x_1 = 2 \), \( y_1 = 25 \), and \( m = 25 \ln(5) \):
\[ y - 25 = 25 \ln(5) (x - 2) \]
5. **Simplify the equation if necessary**:
\[ y = 25 \ln(5) (x - 2) + 25 \]
The tangent line to the graph of \( f(x) = 5^x \) at \( x = 2 \) is thus:
\[ y = 25 \ln(5) (x - 2) + 25 \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6ce675ac-583d-450e-8ae1-ad1f22609723%2Fe9c46e7d-492e-4b54-91fa-26e5dc755fa8%2Fd1xqg5m_reoriented.jpeg&w=3840&q=75)
Transcribed Image Text:## Determining an Equation for the Tangent Line
### Problem Statement:
Determine an equation for the tangent line to the graph at the given value:
\[ f(x) = 5^x, \, x = 2 \]
### Explanation:
To find the equation of the tangent line to the graph of the function \( f(x) = 5^x \) at the point where \( x = 2 \), follow these steps:
1. **Evaluate the function at \( x = 2 \)**:
\[ f(2) = 5^2 = 25 \]
2. **Find the derivative of the function**:
Using the chain rule for the derivative of an exponential function, \( f(x) = 5^x \), the derivative is:
\[ f'(x) = 5^x \ln(5) \]
3. **Evaluate the derivative at \( x = 2 \)**:
\[ f'(2) = 5^2 \ln(5) = 25 \ln(5) \]
4. **Write the equation of the tangent line**:
The equation of a line in point-slope form is given by:
\[ y - y_1 = m(x - x_1) \]
where \( m \) is the slope and \( (x_1, y_1) \) is the point of tangency.
In this case, \( x_1 = 2 \), \( y_1 = 25 \), and \( m = 25 \ln(5) \):
\[ y - 25 = 25 \ln(5) (x - 2) \]
5. **Simplify the equation if necessary**:
\[ y = 25 \ln(5) (x - 2) + 25 \]
The tangent line to the graph of \( f(x) = 5^x \) at \( x = 2 \) is thus:
\[ y = 25 \ln(5) (x - 2) + 25 \]
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