a) To determine where \( f(x) = x^3 – 3x^2 + 4 \) is decreasing:
A function is decreasing when its derivative is negative.
Differentiate \( f(x) \):
\( f'(x) = 3 \times x^2 – 6x = 3x(x – 2) \)

Find critical points by setting \( f'(x) = 0 \):
\( 3x(x – 2) = 0 \)
\( x = 0 \) or \( x = 2 \)

Determine the sign of \( f'(x) \) in the intervals \( (-\infty, 0) \), \( (0, 2) \), and \( (2, \infty) \):
For \( x < 0 \), e.g., \( x = -1 \):
\( f'(-1) = 3 \times (-1) \times (-1 – 2) = 3 \times (-1) \times (-3) = 9 \) (positive)
For \( 0 < x < 2 \), e.g., \( x = 1 \):
\( f'(1) = 3 \times 1 \times (1 – 2) = 3 \times 1 \times (-1) = -3 \) (negative)
For \( x > 2 \), e.g., \( x = 3 \):
\( f'(3) = 3 \times 3 \times (3 – 2) = 3 \times 3 \times 1 = 9 \) (positive)

\( f(x) \) is decreasing where \( f'(x) < 0 \):
\( x \in (0, 2) \) or \( 0 < x < 2 \) [4]

b) To find the point of inflection \( P \):
A point of inflection occurs where the second derivative is zero and changes sign.
First derivative: \( f'(x) = 3x^2 – 6x \)
Second derivative:
\( f”(x) = 6x – 6 \)

Set \( f”(x) = 0 \):
\( 6x – 6 = 0 \)
\( x = 1 \)

Check for sign change in \( f”(x) \):
For \( x < 1 \), e.g., \( x = 0 \):
\( f”(0) = 6 \times 0 – 6 = -6 \) (negative)
For \( x > 1 \), e.g., \( x = 2 \):
\( f”(2) = 6 \times 2 – 6 = 6 \) (positive)

\( f”(x) \) changes sign at \( x = 1 \), confirming a point of inflection.
Find the y-coordinate at \( x = 1 \):
\( f(1) = 1^3 – 3 \times 1^2 + 4 = 1 – 3 + 4 = 2 \)

Coordinates of \( P \):
\( (1, 2) \) [3]