Question
Find an inflection point for the function \(f(x)=2x(x+4)^{3}\)
(A) (0, 0)
(B) (-4, 0)
(C) (-2, 4)
(D) (4, 0)
▶️ Answer/Explanation
First derivative: \( f'(x) = 2 (x + 4)^3 + 6x (x + 4)^2 \).
Second derivative: \( f”(x) = 12 (x + 4)(2x + 4) \).
Set \( f”(x) = 0 \): \( x = -4 \) or \( x = -2 \).
At \( x = -4 \), concavity changes (\( f”(-5) > 0 \), \( f”(-3) < 0 \)); \( f(-4) = 0 \), so point is \((-4, 0)\).
At \( x = -2 \), no change in concavity.
✅ Answer: B) (-4, 0)
Question
If \(f(x)=\left | x^{2} -4\right |\), which of the following statements about f is/are true?
I. f is continuous on the interval (-∞, +∞).
II. f has points of inflection at x = ±2.
III. f has a relative maximum at (0, 4).
(A) I only
(B) II only
(C) III only
(D) I and III
▶️ Answer/Explanation
\( f(x) = x^2 – 4 \) for \( x \leq -2 \) or \( x \geq 2 \), \( f(x) = -x^2 + 4 \) for \( -2 < x < 2 \).
I. Continuity: Limits at \( x = \pm 2 \) match \( f(\pm 2) = 0 \), so \( f \) is continuous everywhere.
II. Inflection points: \( f”(x) = 2 \) or \( -2 \) (no zeros), and \( f”(x) \) undefined at \( x = \pm 2 \) (corners), so no inflection at \( x = \pm 2 \).
III. Relative maximum: \( f(0) = 4 \), \( f(1) = 3 \), \( f(-1) = 3 \), minima at \( x = \pm 2 \) (0), so \( (0, 4) \) is a maximum.
Thus, I and III are true.
✅ Answer: D) I and III
Question
The derivative of
\(f(x)=\frac{x^{4}}{3}-\frac{x^{5}}{5}\) attains its maximum value at x =
(A) –1 (B) 0 (C) 1 (D)\(-\frac{1}{8} \) (E)\(-\frac{1}{2}\)
▶️ Answer/Explanation
The first derivative is: \[ f'(x) = \frac{4x^3}{3} – x^4 \]
The second derivative is: \[ f”(x) = 4x^2 – 4x^3 = 4x^2(1 – x) \]
Critical points occur where \( f”(x) = 0 \): \[ x = 0 \text{ and } x = 1 \]
Concavity test:
| Interval | Test Value | Sign of f”(x) | Concavity |
|---|---|---|---|
| x < 0 | x = -1 | f”(-1) = 4(1) – 4(-1) = 8 > 0 | Up |
| 0 < x < 1 | x = 0.5 | f”(0.5) = 4(0.25) – 4(0.125) = 0.5 > 0 | Up |
| x > 1 | x = 2 | f”(2) = 4(4) – 4(8) = -16 < 0 | Down |
The maximum of \( f'(x) \) occurs at \( x = 1 \) where the concavity changes from upward to downward.
✅ Correct Answer: C) 1
Question
An equation of the line tangent to \(y=x^{3}+3x^{2}+2\) at its point of inflection is
(A)y=-6x-6 (B)y=-3x+1 (C)y=2x+10 (D)y=3x-1 (E)y=4x+1
▶️ Answer/Explanation
Finding the point of inflection:
First derivative: \[ y’ = 3x^2 + 6x \]
Second derivative: \[ y” = 6x + 6 \]
Set \( y” = 0 \) to find inflection point: \[ 6x + 6 = 0 \] \[ x = -1 \]
Concavity analysis:
| Interval | Test Value | Sign of y” | Concavity |
|---|---|---|---|
| x < -1 | x = -2 | y”(-2) = -6 < 0 | Down |
| x > -1 | x = 0 | y”(0) = 6 > 0 | Up |
Finding the tangent line:
1. Point coordinates at x = -1: \[ y(-1) = (-1)^3 + 3(-1)^2 + 2 = -1 + 3 + 2 = 4 \] Point: (-1, 4)
2. Slope at x = -1: \[ y'(-1) = 3(-1)^2 + 6(-1) = 3 – 6 = -3 \]
3. Tangent line equation: \[ y – 4 = -3(x – (-1)) \] \[ y = -3x – 3 + 4 \] \[ y = -3x + 1 \]
✅ Correct Answer: B) y = -3x + 1