Let's go step by step for the function: \[ f(x) = x^4 - 3x^3 \] --- ### (A) Critical values Critical values occur where \( f'(x) = 0 \) or is undefined. First, find the derivative: \[ f'(x) = 4x^3 - 9x^2 \] Set \( f'(x) = 0 \): \[ 4x^3 - 9x^2 = 0 \\ x^2(4x - 9) = 0 \\ x^2 = 0 \implies x = 0 \\ 4x - 9 = 0 \implies x = \frac{9}{4} \] **Critical values:** \( x = 0, \frac{9}{4} \) --- ### (B) Where is \( f(x) \) increasing? Test intervals determined by critical points: \( (-\infty, 0), (0, \frac{9}{4}), (\frac{9}{4}, \infty) \). Test each interval in \( f'(x) = 4x^3 - 9x^2 \): - For \( x < 0 \) (e.g., \( x = -1 \)): \( 4(-1)^3 - 9(-1)^2 = -4 - 9 = -13 \) (Negative, so **decreasing**) - For \( 0 < x < \frac{9}{4} \) (e.g., \( x = 1 \)): \( 4(1)^3 - 9(1)^2 = 4 - 9 = -5 \) (Negative, so **decreasing**) - For \( x > \frac{9}{4} \) (e.g., \( x = 3 \)): \( 4(3)^3 - 9(3)^2 = 4(27) - 9(9) = 108 - 81 = 27 \) (Positive, so **increasing**) **Increasing:** \( \left(\frac{9}{4}, \infty\right) \) --- ### (C) Where is \( f(x) \) decreasing? From above: **Decreasing:** \( (-\infty, 0) \cup (0, \frac{9}{4}) \) --- ### (D) Local maxima Check sign change of \( f'(x) \): - At \( x = 0 \): \( f'(x) \) negative on both sides (no max/min). - At \( x = \frac{9}{4} \): \( f'(x) \) changes from negative to positive (**local minimum**). So, **no local maxima**. **Local maxima at \( x = \) -1000** --- ### (E) Local minima From above, at \( x = \frac{9}{4} \), \( f'(x) \) changes from negative to positive (**local minimum**). **Local minima at \( x = \frac{9}{4} \)** --- ### (F) Concave up Second derivative: \[ f''(x) = 12x^2 - 18x = 6x(2x - 3) \] Set \( f''(x) = 0 \): \[ 6x(2x - 3) = 0 \implies x = 0, \frac{3}{2} \] Test intervals: - \( x < 0 \) (e.g., \( x = -1 \)): \( 6(-1)(2(-1) - 3) = 6(-1)(-2 - 3) = 6(-1)(-5) = 30 \) (Positive, concave up) - \( 0 < x < \frac{3}{2} \) (e.g., \( x = 1 \)): \( 6(1)(2 - 3) = 6(1)(-1) = -6 \) (Negative, concave down) - \( x > \frac{3}{2} \) (e.g., \( x = 2 \)): \( 6(2)(4 - 3) = 6(2)(1) = 12 \) (Positive, concave up) **Concave up:** \( (-\infty, 0) \cup \left(\frac{3}{2}, \infty\right) \) --- ### (G) Concave down **Concave down:** \( (0, \frac{3}{2}) \) --- ### (H) Inflection points Inflection at \( x = 0 \) and \( x = \frac{3}{2} \), where \( f''(x) \) changes sign. **Inflection points:** \( x = 0, \frac{3}{2} \) --- ### (I) Graph complete Enter "1" when done. --- ## **Final Answers** **(A) Critical value(s):** `0,9/4` **(B) Increasing:** `(9/4, infinity)` **(C) Decreasing:** `(-infinity,0) U (0,9/4)` **(D) Local maxima at x =** `-1000` **(E) Local minima at x =** `9/4` **(F) Concave up:** `(-infinity,0) U (3/2,infinity)` **(G) Concave down:** `(0,3/2)` **(H) Inflection point(s) at x =** `0,3/2` **(I) Graph Complete:** `1` --- If you need the graph or more explanation, let me know!

| Question | Final Answer | |--------------------------------------------------------|-------------------------------------------------| | (A) Critical value(s) | 0, 9/4 | | (B) Where is \(f(z)\) increasing | (9/4, ∞) | | (C) Where is \(f(z)\) decreasing | \(-\infty, 0) \cup (0, 9/4)\) | | (D) Local maxima at \(z\) | -1000 (none) | | (E) Local minima at \(z\) | 9/4 | | (F) Concave up | \(-\infty, 0) \cup (3/2, \infty)\) | | (G) Concave down | (0, 3/2) | | (H) Inflection point(s) at \(z\) | 0, 3/2 |