The function is <strong>DECREASING</strong> on:

Consider the following for the next two (02) items that follow: A function is defined by f(x)=\begin{vmatrix}x+1&2&3\\ 2&x+4&6\\ 3&6&x+9\end{vmatrix}.

  1. A. [-\frac{28}{3},0]
  2. B. [0,\frac{28}{3}]
  3. C. [0,\frac{50}{3}]
  4. D. [0,\frac{56}{3}]

Correct Answer: A. [-\frac{28}{3},0]

Explanation

Expanding the determinant: f(x) = (x+1)[(x+4)(x+9)-36] - 2[2(x+9)-18] + 3[12-3(x+4)]. Simplifying gives f(x) = x^3 + 14x^2. The derivative is f'(x) = 3x^2 + 28x = x(3x+28). For the function to be decreasing, f'(x) \leq 0 \implies x(3x+28) \leq 0. The roots are 0 and -\frac{28}{3}, so the interval is [-\frac{28}{3}, 0].

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