1. គេមានអនុគមន៍ y = f(x) = x³ + 3x² - 2  មានក្រាប(C) ដំណាង ។
ក. រកដែនកំណត់ និង គណនាលិមីតចុងដែនៃអនុគមន៍ f ។
ខ. គណនាដេរីវេ f'(x) រួច គូសតារាងសញ្ញា ដេរីវេ។ និង ចំណុចបរមាធៀបនឹងក្រាប ។
គ. រកចំណុចរបត់ ។ និង សមីការបន្ទាត់ ។
ឃ. គូសតារាងអថេរភាពនៃ f(x) ។ គណនា f(-3) , f(1) ។
ង. សង់ក្រាប(C) និង សង់បន្ទាត់ក្នុងតម្រុយតែមួយ ។
Please tell me

For this problem, we are dealing with the function \( y = f(x) = x^3 + 3x^2 - 2 \). We will go through each part step by step:

### ក. ដែនកំណត់ និង លីមីតចុងដែន
**ដែនកំណត់ (Domain):**
The function \( f(x) = x^3 + 3x^2 - 2 \) is a polynomial, and polynomials are defined for all real values of \( x \). Therefore, the domain is:
\[ D(f) = (-\infty, +\infty) \]

**លីមីតចុងដែន (Limits at infinity):**

To find the limits as \( x \) approaches infinity and negative infinity:

\[ \lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} (x^3 + 3x^2 - 2) \]

Since the highest degree term \( x^3 \) dominates as \( x \) becomes very large, the limit is:
\[ \lim_{x \to +\infty} f(x) = +\infty \]

Similarly, for \( x \) approaching negative infinity:

\[ \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} (x^3 + 3x^2 - 2) \]

Again, the highest degree term \( x^3 \) dominates, but since \( x^3 \) becomes very negative as \( x \) becomes very large negatively:
\[ \lim_{x \to -\infty} f(x) = -\infty \]

### ខ. ដេរីវេ f'(x) និង តារាងសញ្ញា
**Finding the derivative \( f'(x) \):**

\[ f(x) = x^3 + 3x^2 - 2 \]

The derivative \( f'(x) \) is:

\[ f'(x) = 3x^2 + 6x \]

**តារាងសញ្ញា (Sign table for \( f'(x) \)):**

To understand the behavior of the function, we find the critical points by setting the derivative equal to zero:

\[ f'(x) = 3x^2 + 6x = 3x(x + 2) = 0 \]

This gives us the critical points:

\[ x = 0 \quad \text{or} \quad x = -2 \]

**Sign of \( f'(x) \):**

- For \( x < -2 \), \( f'(x) > 0 \) (positive)
- For \( -2 < x < 0 \), \( f'(x) < 0 \) (negative)
- For \( x > 0 \), \( f'(x) > 0 \) (positive)

**Conclusion:**
- \( x = -2 \) is a local maximum.
- \( x = 0 \) is a local minimum.

### គ. ចំណុចរបត់ និង សមីការបន្ទាត់
**Concavity and Point of Inflection:**

To find concavity, we compute the second derivative \( f''(x) \):

\[ f''(x) = 6x + 6 \]

Set \( f''(x) = 0 \) to find the point of inflection:

\[ 6x + 6 = 0 \Rightarrow x = -1 \]

At \( x = -1 \), there is a point of inflection. The concavity changes at this point.

**Point of Inflection:**

To find the coordinates of the point of inflection:

\[ f(-1) = (-1)^3 + 3(-1)^2 - 2 = -1 + 3 - 2 = 0 \]

The point of inflection is \( (-1, 0) \).

**Tangent Line Equation at Point of Inflection:**

The slope of the tangent line at the point \( (-1, 0) \) is the derivative evaluated at \( x = -1 \):

\[ f'(-1) = 3(-1)^2 + 6(-1) = 3 - 6 = -3 \]

So, the equation of the tangent line is:

\[ y - 0 = -3(x + 1) \]
\[ y = -3x - 3 \]

### ឃ. តារាងអថេរភាព និង f(-3) , f(1)
**Monotonicity:**

Using the sign of \( f'(x) \):

- \( x < -2 \): \( f(x) \) is increasing.
- \( -2 < x < 0 \): \( f(x) \) is decreasing.
- \( x > 0 \): \( f(x) \) is increasing.

**Calculating \( f(-3) \) and \( f(1) \):**

\[ f(-3) = (-3)^3 + 3(-3)^2 - 2 = -27 + 27 - 2 = -2 \]

\[ f(1) = (1)^3 + 3(1)^2 - 2 = 1 + 3 - 2 = 2 \]

### ង. សង់ក្រាប(C) និង សង់បន្ទាត់ក្នុងតម្រុយ
To sketch the graph \( y = f(x) \), consider the key points and behavior discussed:
- The points \( (-2, f(-2)) \), \( (0, f(0)) \), and \( (-1, 0) \) are crucial.
- Draw the graph with concave changes at \( x = -1 \) and tangent lines at relevant points.

If you need further details or visual graphs, I can provide them.

សិក្សារកដែនកំណត់, ដេរីវេ និងលីមីតចុងដែននៃអនុគមន៍ f(x) = x³ + 3x² - 2 និងសិក្សាតារាងសញ្ញា, ចំណុចបរមាធៀប និងដែនបន្ទាត់នៅក្នុងតម្រុយតែមួយ។ សិក្សានៅក្នុងវិគីភីឌា។

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