Question

If $\Delta \left(x\right)=\left|\begin{array}{ccc}x& 4& 4\\ 9& x& 4\\ 9& 9& x\end{array}\right|,$ then which of the following is (are) CORRECT?

• A. $\Delta \left(x\right)$ is increasing in $(-\infty ,-6)\cup (6,\infty )$
• B. $\Delta \left(x\right)$ is decreasing in $(-6,6)$
• C. $\Delta \left(x\right)$ has a local minimum at $x=6$
• D. $\Delta \left(x\right)$ has a local maximum at $x=6$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\Delta \left(x\right)$ is increasing in $(-\infty ,-6)\cup (6,\infty )$
B $\Delta \left(x\right)$ is decreasing in $(-6,6)$
C $\Delta \left(x\right)$ has a local minimum at $x=6$

$\Delta \left(x\right)=\left|\begin{array}{ccc}x& 4& 4\\ 9& x& 4\\ 9& 9& x\end{array}\right|$
$\Rightarrow {\Delta }^{\prime }\left(x\right)=\left|\begin{array}{ccc}1& 4& 4\\ 0& x& 4\\ 0& 9& x\end{array}\right|+\left|\begin{array}{ccc}x& 0& 4\\ 9& 1& 4\\ 9& 0& x\end{array}\right|+\left|\begin{array}{ccc}x& 4& 0\\ 9& x& 0\\ 9& 9& 1\end{array}\right|$

$\Rightarrow {\Delta }^{\prime }\left(x\right)=3(x^2-36)$
$\Rightarrow {\Delta }^{\prime }\left(x\right)=0\Rightarrow x=\pm 6$


$\Rightarrow \Delta \left(x\right)$ is increasing in $(-\infty ,-6)\cup (6,\infty )$
and $\Delta \left(x\right)$ is decreasing in $(-6,6)$
$\Rightarrow \Delta \left(x\right)$ has a local minimum at $x=6$

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