Question
For what value of k will \(x+\frac{k}{x}\) have a relative maximum at x = −2?
(A) –4 (B) –2 (C) 2 (D) 4 (E) None of these
▶️ Answer/Explanation
Step 1: Find the first derivative
Given \( f(x) = x + \frac{k}{x} \), the first derivative is:
\[ f'(x) = 1 – \frac{k}{x^2} \]
Step 2: Set derivative to zero at x = -2
\[ f'(-2) = 1 – \frac{k}{(-2)^2} = 1 – \frac{k}{4} = 0 \] \[ \frac{k}{4} = 1 \] \[ k = 4 \]
Step 3: Verify it’s a maximum
Second derivative:
\[ f”(x) = \frac{2k}{x^3} \]
At \( x = -2 \) with \( k = 4 \):
\[ f”(-2) = \frac{8}{-8} = -1 < 0 \]
Since \( f”(-2) < 0 \), this confirms a relative maximum at \( x = -2 \).
Conclusion:
The value \( k = 4 \) satisfies the condition for a relative maximum at \( x = -2 \).
✅ Answer: D) 4
Question
At what values of x does \(f(x)=3x^5+5x^3+15\) have a relative maximum?
(A) –1 only (B) 0 only (C) 1 only (D) –1 and 1 only (E) –1, 0 and 1
▶️ Answer/Explanation
1. Find critical points:
\( f'(x) = 15x^4 – 15x^2 = 15x^2(x^2 – 1) \)
Critical points at \( x = -1, 0, 1 \)
2. Test for maximum:
Only \( x = -1 \) changes from \( f’ > 0 \) to \( f’ < 0 \)
Conclusion: Relative maximum only at \( x = -1 \)
✅ Answer: A) -1 only
Question
A polynomial p (x ) has a relative maximum at (−2, 4) , a relative minimum at ( 1,1 ), a relative maximum at ( 5,7) and no other critical points. How many zeros does p ( x) have?
(A) One (B) Two (C) Three (D) Four (E) Five
▶️ Answer/Explanation
Key Observations:
- The polynomial must be at least 4th degree (3 critical points)
- Behavior analysis:
- As x→-∞: p(x)→-∞ (odd degree behavior)
- At (-2,4): Maximum – curve turns downward
- At (1,1): Minimum – curve turns upward
- At (5,7): Maximum – curve turns downward
- As x→∞: p(x)→-∞ (odd degree behavior)
Zero crossings:
- From -∞ to (-2,4): Crosses x-axis once (y changes from -∞ to +4)
- Between (-2,4) and (5,7): Crosses x-axis once (y decreases from 4 to 1 then increases to 7)
- No additional crossings after (5,7) as y decreases from 7 to -∞
✅ Answer: B) Two (exactly 2 zeros)
Question
The derivative of f is \(x^{4}(x-2)(x+3)\) . At how many points will the graph of f have a relative maximum?
(A) None (B) One (C) Two (D) Three (E) Four
▶️ Answer/Explanation
Critical Points:
\( f'(x) = x^4(x-2)(x+3) = 0 \) at \( x = -3, 0, 2 \)
Sign Analysis:
| Interval | Sign of \( f'(x)\) | Behavior |
|---|---|---|
| x < -3 | + | Increasing |
| -3 < x < 0 | – | Decreasing |
| 0 < x < 2 | – | Decreasing |
| x > 2 | + | Increasing |
Relative Maximum:
Only at \( x = -3 \) (\(f’ \)changes from + to -)
At \( x = 0 \): No extremum (even multiplicity)
At \( x = 2 \): Relative minimum (\(f’ \)changes from – to +)
✅ Answer: B) One