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If $f(x)=3 x^{2}+6 x-9$, then
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The correct answer is:
$f(x)$ is decreasing in $(-\infty,-1)$
Given $f(x)=3 x^{2}+6 x-9$ On differentiating w.r.t. $x$, we get $f^{\prime}(x)=6 x+6$
$\begin{aligned} & f^{\prime}(x) < 0 \\ \Rightarrow & 6 x+6 < 0 \end{aligned}$
$\Rightarrow 6 \mathrm{x} < -6$
$\Rightarrow \mathrm{x} < -1$
Hence $f(x)$ is decreasing in $(-\infty,-1)$
$\begin{aligned} & f^{\prime}(x) < 0 \\ \Rightarrow & 6 x+6 < 0 \end{aligned}$
$\Rightarrow 6 \mathrm{x} < -6$
$\Rightarrow \mathrm{x} < -1$
Hence $f(x)$ is decreasing in $(-\infty,-1)$
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