The given function is:

$p(x) = -x^3 + 6x^2 - 9x + 12$

We are asked to find the value of $x$ at which the function reaches a relative maximum. To do this, we need to find the critical points of the function by following these steps:

1. Take the derivative of the function.

The first derivative of $p(x)$, denoted as $p'(x)$, will help us find the critical points:

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

$p'(x) = -3x^2 + 12x - 9$

2. Set the derivative equal to zero to find the critical points.

We solve $p'(x) = 0$:

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

To simplify, divide through by $-3$:

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

Factor the quadratic equation:

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

So, $x = 3$ and $x = 1$.

3. Determine which point is a maximum.

To confirm which critical point is a maximum, we take the second derivative of $p(x)$:

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

$p''(x) = -6x + 12$

Now, evaluate the second derivative at $x = 1$ and $x = 3$:

• For $x = 1$:

$p''(1) = -6(1) + 12 = 6$

Since $p''(1) > 0$, $x = 1$ is a minimum point.

• For $x = 3$:

$p''(3) = -6(3) + 12 = -6$

Since $p''(3) < 0$, $x = 3$ is a maximum point.

Conclusion:

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

Would you like more details or have any other questions?

Here are some related questions to deepen your understanding:

1. How do we find critical points for any polynomial function?
2. What is the significance of the second derivative in determining concavity?
3. Can a function have more than one relative maximum or minimum?
4. How does changing the coefficients in a cubic function affect its graph?
5. Why is dividing the equation by $-3$ important when solving the derivative?

Tip: Always check both the first and second derivatives to correctly classify critical points as maxima, minima, or points of inflection.