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 8:** Find the equation of the horizontal tangent line to the curve \( y = -x + \sqrt{x} \).
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*Explanation:*
To solve this problem, we would first take the derivative of the function \( y = -x + \sqrt{x} \) with respect to \( x \) to find the slope of the tangent line at any point \( x \) on the curve. Next, we set the derivative equal to zero to find the point(s) at which the tangent line is horizontal.
1. Start by differentiating the function \( y \):
\[ \frac{dy}{dx} = \frac{d}{dx} \left(-x + \sqrt{x} \right) \]
2. Use the power rule to differentiate the individual terms within the function:
\[ \frac{dy}{dx} = -1 + \frac{1}{2}x^{-1/2} \]
3. Simplify the derivative:
\[ \frac{dy}{dx} = -1 + \frac{1}{2\sqrt{x}} \]
4. Set the derivative equal to zero to find the point(s) where the tangent line is horizontal:
\[ -1 + \frac{1}{2\sqrt{x}} = 0 \]
5. Solve for \( x \):
\[ \frac{1}{2\sqrt{x}} = 1 \]
\[ \frac{1}{2\sqrt{x}} = 1 \]
\[ \sqrt{x} = \frac{1}{2} \]
\[ x = \left( \frac{1}{2} \right)^2 \]
\[ x = \frac{1}{4} \]
6. Substitute \( x = \frac{1}{4} \) back into the original equation to find \( y \):
\[ y = -\left( \frac{1}{4} \right) + \sqrt{ \frac{1}{4} } \]
\[ y = -\frac{1}{4} + \frac{1}{2} \]
\[ y = \frac{1}{4} \]
Therefore, the horizontal tangent line occurs at the point \( \left( \frac{1}{4}, \frac{1}{4} \right) \). The equation of the horizontal tangent line is:
\[ y = \frac{1}{4} \]
---
In summary, for the curve](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F87082b8a-b4f5-455e-a232-6f60990eaa64%2Fb107a808-1f7f-4b1a-b5c6-c6f9ebb5f026%2Fy6bv8duh_processed.png&w=3840&q=75)
Transcribed Image Text:**Question 8:** Find the equation of the horizontal tangent line to the curve \( y = -x + \sqrt{x} \).
---
*Explanation:*
To solve this problem, we would first take the derivative of the function \( y = -x + \sqrt{x} \) with respect to \( x \) to find the slope of the tangent line at any point \( x \) on the curve. Next, we set the derivative equal to zero to find the point(s) at which the tangent line is horizontal.
1. Start by differentiating the function \( y \):
\[ \frac{dy}{dx} = \frac{d}{dx} \left(-x + \sqrt{x} \right) \]
2. Use the power rule to differentiate the individual terms within the function:
\[ \frac{dy}{dx} = -1 + \frac{1}{2}x^{-1/2} \]
3. Simplify the derivative:
\[ \frac{dy}{dx} = -1 + \frac{1}{2\sqrt{x}} \]
4. Set the derivative equal to zero to find the point(s) where the tangent line is horizontal:
\[ -1 + \frac{1}{2\sqrt{x}} = 0 \]
5. Solve for \( x \):
\[ \frac{1}{2\sqrt{x}} = 1 \]
\[ \frac{1}{2\sqrt{x}} = 1 \]
\[ \sqrt{x} = \frac{1}{2} \]
\[ x = \left( \frac{1}{2} \right)^2 \]
\[ x = \frac{1}{4} \]
6. Substitute \( x = \frac{1}{4} \) back into the original equation to find \( y \):
\[ y = -\left( \frac{1}{4} \right) + \sqrt{ \frac{1}{4} } \]
\[ y = -\frac{1}{4} + \frac{1}{2} \]
\[ y = \frac{1}{4} \]
Therefore, the horizontal tangent line occurs at the point \( \left( \frac{1}{4}, \frac{1}{4} \right) \). The equation of the horizontal tangent line is:
\[ y = \frac{1}{4} \]
---
In summary, for the curve
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