1. សិក្សាអថេរភាព និង សង់ក្រាប នៃអនុគមន៍ $y = f(x) = x^3 - 6x^2 + 9x - 8$

To analyze the behavior of the function $y = f(x) = x^3 - 6x^2 + 9x - 8$ and to sketch its graph, we follow these steps:

a. Finding the First Derivative:

The first derivative $f'(x)$ helps us determine the critical points and the intervals of increase and decrease.

$f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 9x - 8)$ $f'(x) = 3x^2 - 12x + 9$

b. Finding the Critical Points:

Set $f'(x) = 0$ to find the critical points.

$3x^2 - 12x + 9 = 0$

Dividing by 3:

$x^2 - 4x + 3 = 0$

Factoring the quadratic equation:

$(x - 3)(x - 1) = 0$

So, the critical points are $x = 3$ and $x = 1$.

c. Determining Intervals of Increase and Decrease:

We examine the sign of $f'(x)$ in the intervals determined by the critical points:

1. For $x < 1$, choose $x = 0$: $f'(0) = 3(0)^2 - 12(0) + 9 = 9 \quad (\text{positive})$ $f(x)$ is increasing in $(-\infty, 1)$.

2. For $1 < x < 3$, choose $x = 2$: $f'(2) = 3(2)^2 - 12(2) + 9 = -3 \quad (\text{negative})$ $f(x)$ is decreasing in $(1, 3)$.

3. For $x > 3$, choose $x = 4$: $f'(4) = 3(4)^2 - 12(4) + 9 = 9 \quad (\text{positive})$ $f(x)$ is increasing in $(3, \infty)$.

So, the function increases on $(-\infty, 1)$ and $(3, \infty)$, and decreases on $(1, 3)$.

d. Finding the Second Derivative:

The second derivative $f''(x)$ helps us determine concavity and points of inflection.

$f''(x) = \frac{d}{dx}(3x^2 - 12x + 9)$ $f''(x) = 6x - 12$

Set $f''(x) = 0$ to find points of inflection:

$6x - 12 = 0$ $x = 2$

• For $x < 2$, $f''(x) < 0$, so the graph is concave down.
• For $x > 2$, $f''(x) > 0$, so the graph is concave up.

e. Sketching the Graph:

With the above information:

• The graph has a local maximum at $x = 1$ and a local minimum at $x = 3$.
• There is a point of inflection at $x = 2$.
• The function increases on $(-\infty, 1)$ and $(3, \infty)$, and decreases on $(1, 3)$.

2. បង្ហាញថា I(-2, 6) ជាផ្ចិតឆ្លុះនៃក្រាបអនុគមន៍

To verify that $I(-2, 6)$ is a reflection point of the graph:

• First, substitute $x = -2$ into the function:

$f(-2) = (-2)^3 - 6(-2)^2 + 9(-2) - 8 = -8 - 24 - 18 - 8 = -58$

However, this result does not match the point $(x, y) = (-2, 6)$, suggesting a mistake.

Let me recheck by finding the correct point to be substituted into the original function:

1. If $I(-2, 6)$ is a reflection point, the equation of the tangent line at this point should match. However, since the substitution did not work, either the reflection point is wrong, or more analysis is needed.

Would you like more details on any step or the sketching?

5 Related Questions:

1. How do you determine the concavity of a function using the second derivative?
2. What is the significance of the point of inflection in a graph?
3. Can you explain the steps to factor quadratic equations?
4. How do you verify if a point lies on the curve of a function?
5. What methods can be used to find the local extrema of a function?

Tip: Always verify the critical points and evaluate the function to ensure they match the conditions needed for the reflection point.