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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![**Problem Statement:**
Find the equation to the tangent line to the curve \( y = \sqrt{x^3 - 2} \) where \( x = 3 \).
**Solution Approach:**
1. **Compute the point on the curve:**
- Substitute \( x = 3 \) into the equation \( y = \sqrt{x^3 - 2} \).
\[
y = \sqrt{3^3 - 2} = \sqrt{27 - 2} = \sqrt{25} = 5
\]
- Therefore, the point on the curve when \( x = 3 \) is \( (3, 5) \).
2. **Find the derivative \( y' \) to get the slope of the tangent line:**
- The given function is \( y = (x^3 - 2)^{1/2} \).
- Differentiate \( y \) with respect to \( x \):
\[
y' = \frac{d}{dx} \left[(x^3 - 2)^{1/2}\right]
\]
Using the chain rule:
\[
y' = \frac{1}{2}(x^3 - 2)^{-1/2} \cdot 3x^2
\]
Simplified:
\[
y' = \frac{3x^2}{2\sqrt{x^3 - 2}}
\]
- Substitute \( x = 3 \) to find the slope at \( x = 3 \):
\[
y'(3) = \frac{3(3)^2}{2\sqrt{3^3 - 2}} = \frac{3 \cdot 9}{2 \cdot 5} = \frac{27}{10}
\]
3. **Write the equation of the tangent line:**
- The slope of the tangent line at \( x = 3 \) is \( \frac{27}{10} \).
- The point of tangency is \( (3, 5) \).
- Use the point-slope form of the equation of a line:
\[
y - y_1 = m(x - x_1)
\]
Where \( (x_1, y_1) =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdb8ba6cb-f074-43c1-b7bc-05cc004f9981%2Fdbda394f-0118-4ea3-bcd4-fe1b89d19320%2Fy7f0x7_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the equation to the tangent line to the curve \( y = \sqrt{x^3 - 2} \) where \( x = 3 \).
**Solution Approach:**
1. **Compute the point on the curve:**
- Substitute \( x = 3 \) into the equation \( y = \sqrt{x^3 - 2} \).
\[
y = \sqrt{3^3 - 2} = \sqrt{27 - 2} = \sqrt{25} = 5
\]
- Therefore, the point on the curve when \( x = 3 \) is \( (3, 5) \).
2. **Find the derivative \( y' \) to get the slope of the tangent line:**
- The given function is \( y = (x^3 - 2)^{1/2} \).
- Differentiate \( y \) with respect to \( x \):
\[
y' = \frac{d}{dx} \left[(x^3 - 2)^{1/2}\right]
\]
Using the chain rule:
\[
y' = \frac{1}{2}(x^3 - 2)^{-1/2} \cdot 3x^2
\]
Simplified:
\[
y' = \frac{3x^2}{2\sqrt{x^3 - 2}}
\]
- Substitute \( x = 3 \) to find the slope at \( x = 3 \):
\[
y'(3) = \frac{3(3)^2}{2\sqrt{3^3 - 2}} = \frac{3 \cdot 9}{2 \cdot 5} = \frac{27}{10}
\]
3. **Write the equation of the tangent line:**
- The slope of the tangent line at \( x = 3 \) is \( \frac{27}{10} \).
- The point of tangency is \( (3, 5) \).
- Use the point-slope form of the equation of a line:
\[
y - y_1 = m(x - x_1)
\]
Where \( (x_1, y_1) =
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