The function provided is $f(x) = x^3 - 6x^2 + 15$. The question asks for the interval where the function is decreasing.

To determine where the function is decreasing, we first need to find the derivative $f'(x)$ and analyze its sign (whether it's positive or negative). The function will be decreasing where the derivative is negative.

Step 1: Find the derivative

We differentiate the given function $f(x) = x^3 - 6x^2 + 15$.

$f'(x) = \frac{d}{dx} \left( x^3 - 6x^2 + 15 \right)$ $f'(x) = 3x^2 - 12x$

Step 2: Find critical points

Set the derivative equal to zero to find critical points:

$3x^2 - 12x = 0$ Factor out $3x$:

$3x(x - 4) = 0$

This gives us two critical points: $x = 0$ and $x = 4$.

Step 3: Analyze the intervals

We now analyze the sign of $f'(x) = 3x(x - 4)$ in the intervals determined by the critical points: $(-\infty, 0)$, $(0, 4)$, and $(4, \infty)$.

• For $x \in (-\infty, 0)$, choose $x = -1$: $f'(-1) = 3(-1)((-1) - 4) = 3(-1)(-5) = 15 > 0$ So, $f'(x) > 0$ in this interval, meaning the function is increasing.

• For $x \in (0, 4)$, choose $x = 2$: $f'(2) = 3(2)((2) - 4) = 3(2)(-2) = -12 < 0$ So, $f'(x) < 0$ in this interval, meaning the function is decreasing.

• For $x \in (4, \infty)$, choose $x = 5$: $f'(5) = 3(5)((5) - 4) = 3(5)(1) = 15 > 0$ So, $f'(x) > 0$ in this interval, meaning the function is increasing.

Step 4: Conclusion

The function $f(x)$ is decreasing in the interval $(0, 4)$.

The correct answer is: (a) $]0, 4[$.

Would you like further details or have any questions?

Here are 5 related questions:

1. How do we determine whether a function is increasing or decreasing?
2. What is the significance of critical points in analyzing a function?
3. Can a function be increasing in one interval and decreasing in another?
4. How do inflection points differ from critical points?
5. How can we classify the behavior of the function at the critical points?

Tip: The sign of the first derivative is key to understanding the behavior (increasing/decreasing) of a function.