The function attains local <strong>MINIMUM</strong> value at:
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}.
- A. x=-\frac{28}{3}
- B. x=-1
- C. x=0 ✓
- D. x=\frac{28}{3}
Correct Answer: C. x=0
Explanation
From the previous question, f'(x) = x(3x+28). Setting f'(x) = 0, the critical points are x=0 and x=-\frac{28}{3}. The second derivative is f''(x) = 6x + 28. Evaluating at x=0, f''(0) = 28 \gt 0, which implies a local minimum. Evaluating at x=-\frac{28}{3}, f''(-\frac{28}{3}) = -56+28 = -28 \lt 0, implying a local maximum.
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