Let $$ f(x)=\frac{1}{3} x^{3}-2 x^{2} $$. 1. Find critical numbers of the function. Enter your answer here $$ \mathrm{x}= $$ Use a comma to separate answers as needed. If the function has no critical numbers, enter none. 2. Find the $$ x $$-coordinate of a relative maximum point if any. Enter your answer here $$ \mathrm{x}= $$. If the function has no relative maximum points, enter none. 3. Find the $$ x $$-coordinate of a relative minimum point if any. Enter your answer here $$ \mathrm{x}= $$ none. 4. Find the $$ x $$-coordinate of an inflection pc any. Enter your answer here $$ \mathrm{x}= $$ . If the function has no inflection points, enter none.

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**1. Critical Numbers**

The derivative is
$$
f'(x)=\frac{d}{dx}\left(\frac{1}{3}x^3-2x^2\right)=x^2-4x.
$$
Set $$ f'(x) = 0 $$:
$$
x^2-4x=0,\quad x(x-4)=0.
$$
Thus, the critical numbers are
$$
x=0 \quad \text{and} \quad x=4.
$$
So, $$ x=0,\, 4 $$.

---

**2. Relative Maximum Point**

To determine the nature of the critical points, we consider the second derivative:
$$
f''(x)=\frac{d}{dx}\left(x^2-4x\right)=2x-4.
$$
Evaluate at $$ x=0 $$:
$$
f''(0)=2(0)-4=-4.
$$
Since $$ f''(0)<0 $$, the function is concave down at $$ x=0 $$, indicating a relative maximum.

So, the $$ x $$-coordinate of the relative maximum point is $$ x=0 $$.

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**3. Relative Minimum Point**

Evaluate the second derivative at $$ x=4 $$:
$$
f''(4)=2(4)-4=8-4=4.
$$
Since $$ f''(4)>0 $$, the function is concave up at $$ x=4 $$, indicating a relative minimum.

Thus, the $$ x $$-coordinate of the relative minimum point is $$ x=4 $$.

---

**4. Inflection Point**

An inflection point occurs where the concavity changes, i.e., where $$ f''(x)=0 $$. Set the second derivative equal to zero:
$$
2x-4=0 \quad \Longrightarrow \quad x=2.
$$

So, the $$ x $$-coordinate of the inflection point is $$ x=2 $$.
**1. Critical Numbers:** $$ x=0, 4 $$ **2. Relative Maximum Point:** $$ x=0 $$ **3. Relative Minimum Point:** $$ x=4 $$ **4. Inflection Point:** $$ x=2 $$

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