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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Concept explainers
Rate of Change
The relation between two quantities which displays how much greater one quantity is than another is called ratio.
Slope
The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:
Question
![**Problem:**
Find all \( x \)-values where \( f(x) = (x^2 + 2x - 3)^4 \) has a horizontal tangent line.
**Explanation:**
To find where the function has a horizontal tangent line, we need to find the derivative \( f'(x) \) and set it to zero, as horizontal tangent lines occur where the slope of the function is zero.
1. **Differentiate the Function:**
- First, let \( u = x^2 + 2x - 3 \), so \( f(x) = u^4 \).
- The derivative of \( f(x) \) with respect to \( x \) using the chain rule is \( f'(x) = 4u^3 \cdot \frac{du}{dx} \).
2. **Find \( \frac{du}{dx} \):**
- \( \frac{du}{dx} = 2x + 2 \).
3. **Substitute Back:**
- \( f'(x) = 4(x^2 + 2x - 3)^3 \cdot (2x + 2) \).
4. **Set the Derivative to Zero:**
\[
4(x^2 + 2x - 3)^3 \cdot (2x + 2) = 0
\]
5. **Solve for \( x \):**
- Either \( (x^2 + 2x - 3) = 0 \) or \( (2x + 2) = 0 \).
- Solving \( x^2 + 2x - 3 = 0 \) gives the roots using the quadratic formula:
\[
x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1}
\]
\[
x = \frac{-2 \pm \sqrt{16}}{2}
\]
\[
x = \frac{-2 \pm 4}{2}
\]
\( x = 1 \) or \( x = -3 \).
- Solving \( 2x + 2 = 0 \) gives:
\[
x = -1
\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F94ff4fec-8c60-4ff2-9b27-1ba17c465d2d%2Fa1741f09-3214-4f92-a29a-fbb840d19be4%2Fiyzquuj_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem:**
Find all \( x \)-values where \( f(x) = (x^2 + 2x - 3)^4 \) has a horizontal tangent line.
**Explanation:**
To find where the function has a horizontal tangent line, we need to find the derivative \( f'(x) \) and set it to zero, as horizontal tangent lines occur where the slope of the function is zero.
1. **Differentiate the Function:**
- First, let \( u = x^2 + 2x - 3 \), so \( f(x) = u^4 \).
- The derivative of \( f(x) \) with respect to \( x \) using the chain rule is \( f'(x) = 4u^3 \cdot \frac{du}{dx} \).
2. **Find \( \frac{du}{dx} \):**
- \( \frac{du}{dx} = 2x + 2 \).
3. **Substitute Back:**
- \( f'(x) = 4(x^2 + 2x - 3)^3 \cdot (2x + 2) \).
4. **Set the Derivative to Zero:**
\[
4(x^2 + 2x - 3)^3 \cdot (2x + 2) = 0
\]
5. **Solve for \( x \):**
- Either \( (x^2 + 2x - 3) = 0 \) or \( (2x + 2) = 0 \).
- Solving \( x^2 + 2x - 3 = 0 \) gives the roots using the quadratic formula:
\[
x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1}
\]
\[
x = \frac{-2 \pm \sqrt{16}}{2}
\]
\[
x = \frac{-2 \pm 4}{2}
\]
\( x = 1 \) or \( x = -3 \).
- Solving \( 2x + 2 = 0 \) gives:
\[
x = -1
\
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