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...
Related questions
Question
![**Calculating the Equation of the Tangent Line**
*Problem Statement:*
Assume that functions \(f\) and \(g\) are differentiable with the following values:
- \(f(4) = -2\)
- \(f'(4) = 3\)
- \(g(4) = -4\)
- \(g'(4) = -2\)
Find an equation of the line tangent to the graph of \(F(x) = f(x)g(x)\) at \(x = 4\).
*Solution:*
To find the equation of the tangent line, we need to determine the derivative \(F'(x)\) using the product rule:
\[
F(x) = f(x) \cdot g(x)
\]
The product rule states that:
\[
F'(x) = f'(x)g(x) + f(x)g'(x)
\]
Substituting the given values at \(x = 4\):
\[
F'(4) = f'(4)g(4) + f(4)g'(4) = (3 \cdot -4) + (-2 \cdot -2) = -12 + 4 = -8
\]
The point on the curve at \(x = 4\) is:
- \(F(4) = f(4) \cdot g(4) = -2 \cdot -4 = 8\)
The equation of the tangent line in point-slope form is:
\[
y - F(4) = F'(4)(x - 4)
\]
Substituting the known values:
\[
y - 8 = -8(x - 4)
\]
Simplifying:
\[
y = -8x + 32 + 8 = -8x + 40
\]
*Conclusion:*
The equation of the tangent line is:
\[
y = -8x + 40
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F10b1f275-2c41-4985-b23a-aa3382c0d493%2Fc792d54c-f9a8-460e-bd05-6593756dc91e%2F3py27_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Calculating the Equation of the Tangent Line**
*Problem Statement:*
Assume that functions \(f\) and \(g\) are differentiable with the following values:
- \(f(4) = -2\)
- \(f'(4) = 3\)
- \(g(4) = -4\)
- \(g'(4) = -2\)
Find an equation of the line tangent to the graph of \(F(x) = f(x)g(x)\) at \(x = 4\).
*Solution:*
To find the equation of the tangent line, we need to determine the derivative \(F'(x)\) using the product rule:
\[
F(x) = f(x) \cdot g(x)
\]
The product rule states that:
\[
F'(x) = f'(x)g(x) + f(x)g'(x)
\]
Substituting the given values at \(x = 4\):
\[
F'(4) = f'(4)g(4) + f(4)g'(4) = (3 \cdot -4) + (-2 \cdot -2) = -12 + 4 = -8
\]
The point on the curve at \(x = 4\) is:
- \(F(4) = f(4) \cdot g(4) = -2 \cdot -4 = 8\)
The equation of the tangent line in point-slope form is:
\[
y - F(4) = F'(4)(x - 4)
\]
Substituting the known values:
\[
y - 8 = -8(x - 4)
\]
Simplifying:
\[
y = -8x + 32 + 8 = -8x + 40
\]
*Conclusion:*
The equation of the tangent line is:
\[
y = -8x + 40
\]
Expert Solution
Step 1
Given :
Assume that the functions and are differentiable with and .
To find :
The equation of the line tangent to the graph of at .
Formula used :
(1) Product rule of differentiation is .
(2) The equation of tangent line with slope at the point is .
Step by step
Solved in 3 steps
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