12. A function f(x) is differentiable at a = 2. Which of the following is guaranteed to equal f'(2)? (A) I only (D) I and III only I. II. III. lim #12 lim #12 f(x)-f(2) x-2 (B) II only f(2)-f(x) 2 x lim h→0 f(2+h)-f(2) h (C) I and II only (E) I, II, and III
12. A function f(x) is differentiable at a = 2. Which of the following is guaranteed to equal f'(2)? (A) I only (D) I and III only I. II. III. lim #12 lim #12 f(x)-f(2) x-2 (B) II only f(2)-f(x) 2 x lim h→0 f(2+h)-f(2) h (C) I and II only (E) I, II, and III
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 12: Differentiability and Limits**
**Problem Statement:**
A function \( f(x) \) is differentiable at \( x = 2 \). Which of the following is guaranteed to equal \( f'(2) \)?
**Options:**
I. \( \lim_{x \to 2} \frac{f(x) - f(2)}{x - 2} \)
II. \( \lim_{x \to 2} \frac{f(2) - f(x)}{2 - x} \)
III. \( \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h} \)
**Answer Choices:**
- (A) I only
- (B) II only
- (C) I and II only
- (D) I and III only
- (E) I, II, and III
**Explanation:**
To determine which of the given limits are guaranteed to equal \( f'(2) \), we need to recall the definition of the derivative. The derivative of a function \( f \) at \( x = 2 \) is given by:
\[ f'(2) = \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h} \]
This is expressed in option III, so III is guaranteed to equal \( f'(2) \).
For differentiability, another equivalent form is:
\[ f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \]
When \( a = 2 \), this is:
\[ f'(2) = \lim_{x \to 2} \frac{f(x) - f(2)}{x - 2} \]
This corresponds to option I. Hence, I is also guaranteed to equal \( f'(2) \).
For option II, by examining the expression:
\[ \lim_{x \to 2} \frac{f(2) - f(x)}{2 - x} \]
we can simplify by noting that \( 2 - x \) tends to zero as \( x \) tends to 2, just as \( x - 2 \) does. Therefore, this limit is equivalent to limit I (since multiplying the numerator and denominator by -1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0c4c60bc-d226-400e-a532-f256a6d25f22%2Ff978d21f-b8e0-444c-8758-628ffcc911de%2F5ugbqum_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 12: Differentiability and Limits**
**Problem Statement:**
A function \( f(x) \) is differentiable at \( x = 2 \). Which of the following is guaranteed to equal \( f'(2) \)?
**Options:**
I. \( \lim_{x \to 2} \frac{f(x) - f(2)}{x - 2} \)
II. \( \lim_{x \to 2} \frac{f(2) - f(x)}{2 - x} \)
III. \( \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h} \)
**Answer Choices:**
- (A) I only
- (B) II only
- (C) I and II only
- (D) I and III only
- (E) I, II, and III
**Explanation:**
To determine which of the given limits are guaranteed to equal \( f'(2) \), we need to recall the definition of the derivative. The derivative of a function \( f \) at \( x = 2 \) is given by:
\[ f'(2) = \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h} \]
This is expressed in option III, so III is guaranteed to equal \( f'(2) \).
For differentiability, another equivalent form is:
\[ f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \]
When \( a = 2 \), this is:
\[ f'(2) = \lim_{x \to 2} \frac{f(x) - f(2)}{x - 2} \]
This corresponds to option I. Hence, I is also guaranteed to equal \( f'(2) \).
For option II, by examining the expression:
\[ \lim_{x \to 2} \frac{f(2) - f(x)}{2 - x} \]
we can simplify by noting that \( 2 - x \) tends to zero as \( x \) tends to 2, just as \( x - 2 \) does. Therefore, this limit is equivalent to limit I (since multiplying the numerator and denominator by -1
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