The problem provides the function $p(x) = -x^3 + 6x^2 - 9x + 12$ and asks for the value of $x$ at which the function reaches a relative maximum (in Indonesian: "mencapai balik maksimum pada saat x").

To solve this, we need to follow these steps:

Step 1: Find the first derivative of the function

To determine the critical points, we first calculate the first derivative $p'(x)$:

$p'(x) = \frac{d}{dx}(-x^3 + 6x^2 - 9x + 12)$ Using standard differentiation rules: $p'(x) = -3x^2 + 12x - 9$

Step 2: Set the derivative equal to zero

To find the critical points, we solve $p'(x) = 0$:

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

We can simplify by dividing the equation by $-3$: $x^2 - 4x + 3 = 0$

Step 3: Solve the quadratic equation

Now, we solve the quadratic equation $x^2 - 4x + 3 = 0$ using the factorization method: $(x - 3)(x - 1) = 0$ Thus, $x = 3$ and $x = 1$.

Step 4: Determine the nature of the critical points

To determine whether these points are maxima or minima, we need to check the second derivative $p''(x)$:

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

• At $x = 1$, $p''(1) = -6(1) + 12 = 6$ (positive), so $x = 1$ is a minimum.
• At $x = 3$, $p''(3) = -6(3) + 12 = -6$ (negative), so $x = 3$ is a maximum.

Final Answer:

The function reaches a relative maximum at $x = 3$.

Would you like more details or explanations?

Here are 5 related questions:

1. How can we determine the relative minima for a function?
2. What is the significance of the second derivative in determining maxima or minima?
3. Can you describe the difference between absolute and relative maxima?
4. How can the first derivative test be used to classify critical points?
5. What happens if the second derivative is zero at a critical point?

Tip: Always check the second derivative when classifying critical points to determine whether the point is a maximum, minimum, or neither.