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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How do you algebraically prove that the two lines are parallel?(Please show work)
![**Problem 19: Calculating Parallel Secant and Tangent Lines**
The problem involves finding the value of \( c \) in the function \( f(x) = \sqrt{x} \) over the interval \([0, 4]\) such that the secant line and the tangent line are parallel.
**Solution:**
\[ f(c) = \sqrt{c}, \quad f(x) = \sqrt{x}, \quad [0, 4] \]
The formula for the slope of the secant line is:
\[ \frac{f(4) - f(0)}{4 - 0} \]
Substitute the values into the equation:
\[ \frac{2 - 0}{4} = \frac{1}{2} \]
The formula for the derivative of \( f(x) = \sqrt{x} \) is:
\[ f'(x) = \frac{1}{2\sqrt{x}} \]
Setting the slope of the tangent line equal to the slope of the secant line:
\[ \frac{1}{2\sqrt{c}} = \frac{1}{2} \]
Solving for \( c \):
\[ \sqrt{c} = 1 \]
\[ c = 1 \]
Thus, when \( c = 1 \), the secant line and the tangent line are parallel.
**Diagram Explanation:**
The diagram in the image is a rectangle with points marked on the upper and lower boundaries, illustrating a secant line connecting the points. The secant line represents the average rate of change of the function over the interval. The setup indicates how the tangent at \( x = c \) matches this slope, emphasizing parallelism between the secant and tangent lines.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F78ee009c-71d4-4c99-a943-0afa7af96a89%2F5185a7b2-9584-43c1-9384-c664ec4a477f%2F0wdw5vh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 19: Calculating Parallel Secant and Tangent Lines**
The problem involves finding the value of \( c \) in the function \( f(x) = \sqrt{x} \) over the interval \([0, 4]\) such that the secant line and the tangent line are parallel.
**Solution:**
\[ f(c) = \sqrt{c}, \quad f(x) = \sqrt{x}, \quad [0, 4] \]
The formula for the slope of the secant line is:
\[ \frac{f(4) - f(0)}{4 - 0} \]
Substitute the values into the equation:
\[ \frac{2 - 0}{4} = \frac{1}{2} \]
The formula for the derivative of \( f(x) = \sqrt{x} \) is:
\[ f'(x) = \frac{1}{2\sqrt{x}} \]
Setting the slope of the tangent line equal to the slope of the secant line:
\[ \frac{1}{2\sqrt{c}} = \frac{1}{2} \]
Solving for \( c \):
\[ \sqrt{c} = 1 \]
\[ c = 1 \]
Thus, when \( c = 1 \), the secant line and the tangent line are parallel.
**Diagram Explanation:**
The diagram in the image is a rectangle with points marked on the upper and lower boundaries, illustrating a secant line connecting the points. The secant line represents the average rate of change of the function over the interval. The setup indicates how the tangent at \( x = c \) matches this slope, emphasizing parallelism between the secant and tangent lines.
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