▶️ Answer/Explanation
Solution
Use the Second Derivative Test:
At \( x = 1 \): \( f'(1) = 0 \), \( f”(1) = -2.0 < 0 \) implies a local maximum.
At \( x = 4 \): \( f'(4) = 0 \), \( f”(4) = 1.2 > 0 \) implies a local minimum.
Evaluate statements:
– I) False: \( x = 1 \) is a maximum.
– II) True: \( x = 1 \) is a maximum.
– III) False: \( x = 4 \) is a minimum.
Answer: B
Final Answer: II only
▶️ Answer/Explanation
Solution
Find the derivative: \( f'(x) = \frac{d}{dx} \left( \frac{x^4}{3} – \frac{x^5}{5} \right) = \frac{4x^3}{3} – x^4 \).
To find where \( f'(x) \) is maximized, compute the second derivative: \( f”(x) = \frac{d}{dx} \left( \frac{4x^3}{3} – x^4 \right) = 4x^2 – 4x^3 = 4x^2(1 – x) \).
Set \( f”(x) = 0 \) to find critical points: \( 4x^2(1 – x) = 0 \) implies \( x = 0 \) or \( x = 1 \).
Analyze the sign of \( f”(x) \):
– For \( x < 0 \), \( 4x^2 > 0 \), \( 1 – x > 0 \), so \( f”(x) > 0 \) (concave up).
– For \( 0 < x < 1 \), \( 4x^2 > 0 \), \( 1 – x > 0 \), so \( f”(x) > 0 \) (concave up).
– For \( x > 1 \), \( 4x^2 > 0 \), \( 1 – x < 0 \), so \( f”(x) < 0 \) (concave down).
At \( x = 1 \), \( f”(x) \) changes from positive to negative, indicating a maximum of \( f'(x) \).
Check \( f'(x) \) at critical points:
– \( f'(0) = 0 \)
– \( f'(1) = \frac{4(1)^3}{3} – (1)^4 = \frac{4}{3} – 1 = \frac{1}{3} \)
Since \( f”(x) < 0 \) at \( x = 1 \), \( f'(x) \) has a maximum at \( x = 1 \).
Answer: C
Final Answer: 1
▶️ Answer/Explanation
Solution
Find inflection: \( f'(x) = 3x^2 + 6x \), \( f”(x) = 6x + 6 \). Set \( f”(x) = 0 \): \( x = -1 \).
Point: \( y = (-1)^3 + 3(-1)^2 + 2 = 4 \), so \( (-1, 4) \).
Slope: \( f'(-1) = 3(-1)^2 + 6(-1) = -3 \).
Tangent: \( y – 4 = -3(x + 1) \), \( y = -3x + 1 \).
Answer: B