▶️ Answer/Explanation
Solution
Tangent line at \( x = 2 \): \( y = f'(2)(x – 2) + f(2) \).
Given \( f'(2) = 4 \), \( f(2) = 1 \), so \( y = 4(x – 2) + 1 \).
Simplify: \( y = 4x – 8 + 1 = 4x – 7 \).
At \( x = 1.9 \): \( f(1.9) \approx 4(1.9) – 7 = 7.6 – 7 = 0.6 \).
Answer: B
Final Answer: 0.6
▶️ Answer/Explanation
Solution
Critical points occur where \( f'(x) = 0 \). Approximate \( f’ \) using differences:
x = 10: \( f'(10) ≈ 2 – 5 = 3 \)
x = 11: \( f'(11) ≈ 3 – 2 = 1 \)
x = 12: \( f'(12) ≈ 6 – 3 = 3 \)
x = 13: \( f'(13) ≈ 5 – 6 = 1 \)
Changes in \( f’ \) suggest critical points between 10-11 (sign change) and 12-13 (sign change), giving two critical points.
For exactly two critical points, \( f'(12) ≠ 0 \) must hold, as \( f'(12) ≈ 3 ≠ 0 \).
Answer: D
Final Answer: \( f'(12) ≠ 0 \)
Question
The graph of f”,the second derivative of the function f, is shown above. Which of the following could be the graph of f ?
A 
B 
C 
D 
▶️Answer/Explanation
Ans:D
▶️ Answer/Explanation
Solution
To determine the correct order of derivatives:
1. At x = -2:
The graph shows a steep increasing slope ⇒ \( f'(-2) \) is large and positive
2. At x = 0:
The graph reaches a local maximum ⇒ \( f'(0) = 0 \) (horizontal tangent)
3. At x = 3:
The graph shows a decreasing slope ⇒ \( f'(3) \) is negative
4. Ordering:
Negative < 0 < Positive ⇒ \( f'(3) < f'(0) < f'(-2) \)
Key observations:
- Steepness determines magnitude of derivative
- Sign indicates increasing/decreasing
- Local extrema have zero derivative
✅ Therefore, the correct answer is D) \( f'(3) < f'(0) < f'(-2) \).