tix)=Dx²-6xs hurizontal tumpant eme fir). f')=D(x2-6xj Find x-value

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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The image contains handwritten notes on finding the x-value for which there is a horizontal tangent line to a given function.

---

**Function:**
\[ f(x) = (x^2 - 6x)^{1/3} \]

**Goal:**
Find the x-value for which there is a horizontal tangent line, i.e., where the derivative \( f'(x) = 0 \).

**Derivative Calculation:**

1. Differentiate the function using the chain rule:
   \[
   f'(x) = \frac{1}{3} (x^2 - 6x)^{-2/3} \cdot (2x - 6)
   \]

2. Rewrite the expression:
   \[
   f'(x) = \frac{1}{3} \cdot 2x (x^2 - 6x)^{-2/3} - 6 (x^2 - 6x)^{-2/3}
   \]

**Next Steps:**
Find the x-value where \( f'(x) = 0 \).

--- 

Note: The visibility of some parts of the equation may be limited, and further steps should involve solving \( f'(x) = 0 \) for the x-value.
Transcribed Image Text:The image contains handwritten notes on finding the x-value for which there is a horizontal tangent line to a given function. --- **Function:** \[ f(x) = (x^2 - 6x)^{1/3} \] **Goal:** Find the x-value for which there is a horizontal tangent line, i.e., where the derivative \( f'(x) = 0 \). **Derivative Calculation:** 1. Differentiate the function using the chain rule: \[ f'(x) = \frac{1}{3} (x^2 - 6x)^{-2/3} \cdot (2x - 6) \] 2. Rewrite the expression: \[ f'(x) = \frac{1}{3} \cdot 2x (x^2 - 6x)^{-2/3} - 6 (x^2 - 6x)^{-2/3} \] **Next Steps:** Find the x-value where \( f'(x) = 0 \). --- Note: The visibility of some parts of the equation may be limited, and further steps should involve solving \( f'(x) = 0 \) for the x-value.
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