▶️ Answer/Explanation
Solution
To find the number of relative extrema for \( g \):
1. Compute the derivative:
- \( g'(x) = \frac{d}{dx}(x^4 + 4x^3) = 4x^3 + 12x^2 = 4x^2(x + 3) \)
2. Find critical points by setting \( g'(x) = 0 \):
- \( 4x^2(x + 3) = 0 \) at \( x = 0 \) or \( x = -3 \)
3. Determine sign changes of \( g'(x) \):
- Test \( (-\infty, -3) \): At \( x = -4 \), \( 4(-4)^2(-1) = -64 < 0 \)
- Test \( (-3, 0) \): At \( x = -1 \), \( 4(-1)^2(2) = 8 > 0 \)
- Test \( (0, \infty) \): At \( x = 1 \), \( 4(1)^2(4) = 16 > 0 \)
- At \( x = -3 \), sign changes from negative to positive (relative minimum).
- At \( x = 0 \), no sign change due to the \( x^2 \) factor (remains positive).
4. Count relative extrema:
- Only one relative extremum at \( x = -3 \).
✅ The correct answer is B) One.
▶️ Answer/Explanation
Solution
To find the relative extrema of \( f \):
1. Find critical points by setting \( f'(x) = 0 \):
- \( f'(x) = (x – 2)(x – 3)^2 (x – 4)^3 = 0 \) at \( x = 2 \), \( x = 3 \), \( x = 4 \)
2. Analyze sign changes of \( f'(x) \) around critical points:
- Test \( (-\infty, 2) \): At \( x = 0 \), \( (-2)(9)(-64) = 1152 > 0 \)
- Test \( (2, 3) \): At \( x = 2.5 \), \( (0.5)(0.25)(-1.5)^3 < 0 \)
- Test \( (3, 4) \): At \( x = 3.5 \), \( (1.5)(0.25)(-0.5)^3 < 0 \)
- Test \( (4, \infty) \): At \( x = 5 \), \( (3)(4)(1) = 12 > 0 \)
3. Determine extrema based on sign changes:
- At \( x = 2 \) (simple root): Sign changes from positive to negative (\( + \) to \( – \)), so a relative maximum.
- At \( x = 3 \) (double root): No sign change (remains negative), so no extremum.
- At \( x = 4 \) (triple root): No sign change (remains negative to positive), so no extremum.
4. Match with options:
- \( f \) has a relative maximum at \( x = 2 \) (I only).
✅ The correct answer is A) I.
▶️ Answer/Explanation
Solution
To find the relative maximum of \( f \):
1. Compute the derivative:
- \( f'(x) = \frac{d}{dx}(x^3 – 3x^2) = 3x^2 – 6x = 3x(x – 2) \)
2. Find critical points by setting \( f'(x) = 0 \):
- \( 3x(x – 2) = 0 \) at \( x = 0 \) or \( x = 2 \)
3. Use the first derivative test to determine the nature of the critical points:
- Test \( (-\infty, 0) \): At \( x = -1 \), \( 3(-1)(-3) = 9 > 0 \)
- Test \( (0, 2) \): At \( x = 1 \), \( 3(1)(-1) = -3 < 0 \)
- Test \( (2, \infty) \): At \( x = 3 \), \( 3(3)(1) = 9 > 0 \)
- At \( x = 0 \), sign changes from positive to negative (\( + \) to \( – \)), so a relative maximum.
- At \( x = 2 \), sign changes from negative to positive (\( – \) to \( + \)), so a relative minimum.
4. The relative maximum occurs at:
- \( x = 0 \)
✅ The correct answer is B) 0.
▶️ Answer/Explanation
Solution
To find the values of \( x \) where \( f \) has a relative maximum:
1. Compute the derivative:
- \( f'(x) = \frac{d}{dx}(3x^5 – 5x^3 + 15) = 15x^4 – 15x^2 = 15x^2(x – 1)(x + 1) \)
2. Find critical points by setting \( f'(x) = 0 \):
- \( 15x^2(x – 1)(x + 1) = 0 \) at \( x = -1 \), \( x = 0 \), \( x = 1 \)
3. Use the first derivative test to determine the nature of the critical points:
- Test \( (-\infty, -1) \): At \( x = -2 \), \( 15(4)(-3)(-1) = 180 > 0 \)
- Test \( (-1, 0) \): At \( x = -0.5 \), \( 15(0.25)(-1.5)(0.5) < 0 \)
- Test \( (0, 1) \): At \( x = 0.5 \), \( 15(0.25)(-0.5)(1.5) < 0 \)
- Test \( (1, \infty) \): At \( x = 2 \), \( 15(4)(1)(3) = 180 > 0 \)
- At \( x = -1 \), sign changes from positive to negative (\( + \) to \( – \)), so a relative maximum.
- At \( x = 0 \), no sign change (remains negative), so no extremum.
- At \( x = 1 \), sign changes from negative to positive (\( – \) to \( + \)), so a relative minimum.
4. The relative maximum occurs at:
- \( x = -1 \)
✅ The correct answer is A) -1 only.