Let's solve the problem step by step for the function \( f(x) = x^3 - 6x^2 + 2 \): --- ### (A) **Critical Values** Critical values are found where \( f'(x) = 0 \) or \( f'(x) \) does not exist. First, find the first derivative: \[ f'(x) = 3x^2 - 12x \] Set \( f'(x) = 0 \): \[ 3x^2 - 12x = 0 \implies 3x(x - 4) = 0 \implies x = 0, \; x = 4 \] **Critical value(s):** \[ \boxed{0, 4} \] --- ### (B) **Where \( f(x) \) is Increasing** Check the sign of \( f'(x) \): - Pick test points in the intervals determined by critical points: \((-\infty, 0)\), \((0, 4)\), \((4, \infty)\). 1. **For \( x < 0 \):** \( f'(x) = 3x^2 - 12x \) For example, at \( x = -1 \): \( f'(-1) = 3(-1)^2 - 12(-1) = 3 + 12 = 15 > 0 \) So, increasing on \( (-\infty, 0) \) 2. **For \( 0 < x < 4 \):** Try \( x = 1 \): \( f'(1) = 3(1)^2 - 12(1) = 3 - 12 = -9 < 0 \) So, decreasing on \( (0, 4) \) 3. **For \( x > 4 \):** Try \( x = 5 \): \( f'(5) = 3(25) - 12(5) = 75 - 60 = 15 > 0 \) So, increasing on \( (4, \infty) \) **Increasing:** \[ \boxed{(-\infty, 0) \cup (4, \infty)} \] --- ### (C) **Where \( f(x) \) is Decreasing** From above, decreasing on \( (0, 4) \). **Decreasing:** \[ \boxed{(0, 4)} \] --- ### (D) **Local Maxima** Use the **First Derivative Test**: - At \( x = 0 \): \( f'(x) \) changes from positive (left) to negative (right) ⇒ **local maximum** - At \( x = 4 \): \( f'(x) \) changes from negative (left) to positive (right) ⇒ **local minimum** **Local maxima at \( x = \):** \[ \boxed{0} \] --- ### (E) **Local Minima** From above, at \( x = 4 \): **Local minima at \( x = \):** \[ \boxed{4} \] --- ### (F) **Concave Up** Find the second derivative: \[ f''(x) = 6x - 12 \] Set \( f''(x) = 0 \): \[ 6x - 12 = 0 \implies x = 2 \] Test intervals: 1. \( x < 2 \): Try \( x = 0 \): \( f''(0) = -12 < 0 \) (concave down) 2. \( x > 2 \): Try \( x = 3 \): \( f''(3) = 18 - 12 = 6 > 0 \) (concave up) **Concave up:** \[ \boxed{(2, \infty)} \] --- ### (G) **Concave Down** From above, concave down on \( (-\infty, 2) \). **Concave down:** \[ \boxed{(-\infty, 2)} \] --- ### (H) **Inflection Points** Inflection points occur where \( f''(x) = 0 \) and the concavity changes. From above, at \( x = 2 \). **Inflection point(s) at \( x = \):** \[ \boxed{2} \] --- ### (I) **Graph Complete** After entering all the above, enter: \[ \boxed{1} \] --- ## **Summary:** | Part | Answer | |--------|---------------------------------| | (A) | 0, 4 | | (B) | (-∞, 0) U (4, ∞) | | (C) | (0, 4) | | (D) | 0 | | (E) | 4 | | (F) | (2, ∞) | | (G) | (-∞, 2) | | (H) | 2 | | (I) | 1 | If you need the actual graph sketch, let me know!

Critical value(s) = 0, 4 Increasing: (-∞, 0) U (4, ∞) Decreasing: (0, 4) Local maxima at = 0 Local minima at = 4 Concave up: (2, ∞) Concave down: (-∞, 2) Inflection point(s) at = 2 Graph Complete: 1

Certainly! Here's a step-by-step guide to sketch the graph of \(f(x) = x^3 - 6x^2 + 2\) based on the information provided: --- ### Step 1: Critical Points - Critical points at \(x=0\) (local max) and \(x=4\) (local min). ### Step 2: Increasing/Decreasing - Increasing on \((-\infty, 0)\) and \((4, \infty)\). - Decreasing on \((0, 4)\). ### Step 3: Concavity and Inflection Point - Concave down on \((-\infty, 2)\). - Concave up on \((2, \infty)\). - Inflection point at \(x=2\). --- ### **Sketching the graph:** 1. **Start from the left:** - As \(x \to -\infty\), \(f(x) \to -\infty\). - Since the function is increasing here, draw the curve rising from the bottom left. 2. **Near \(x=0\):** - At \(x=0\), there's a local maximum. - Calculate \(f(0) = 0^3 - 6 \times 0^2 + 2 = 2\). - So, plot a point at \((0, 2)\) and make it a peak. 3. **Between \(0\) and \(4\):** - The function decreases from the maximum at \(x=0\) down to the minimum at \(x=4\). 4. **At \(x=4\):** - Local minimum with \(f(4) = 4^3 - 6 \times 4^2 + 2 = 64 - 96 + 2 = -30\). - Plot point \((4, -30)\). 5. **Beyond \(x=4\):** - The function increases again to \(+\infty\). 6. **Concavity and Inflection:** - At \(x=2\), the concavity changes. - Find \(f(2) = 8 - 24 + 2 = -14\). - Plot \((2, -14)\), which is an inflection point where the curve changes from concave down to concave up. 7. **Shape summary:** - From \(-\infty\) to 0: increasing and concave down. - From 0 to 2: decreasing and concave down. - At \(x=2\): inflection point. - From 2 to 4: decreasing and concave up. - After 4: increasing and concave up. --- ### **Final sketch tips:** - Draw a smooth curve starting from the bottom left, rising to a maximum at \((0,2)\). - Then, descend through \((2, -14)\), changing concavity at the inflection point. - Continue downward to the minimum at \((4, -30)\). - Then, rise again to the top right, increasing to infinity. --- **Note:** Since I can't draw images directly here, I recommend using graphing tools like Desmos or GeoGebra to visualize the described features based on these critical points, concavity, and inflection points. --- If you'd like, I can help generate a precise plot using an online tool!