Question
The interval in which the function $f(x)=2 x^3+9 x^2+$ $12 x-1$ is decreasing, is
✓
Answer
We have, $f(x)=2 x^3+9 x^2+12 x-1$$
\Rightarrow f^{\prime}(x)=6 x^2+18 x+12
$
For decreasing, $f^{\prime}(x)<0$
$
\begin{array}{ll}
\therefore & 6 x^2+18 x+12<0 \\
\Rightarrow & x^2+3 x+2<0 \Rightarrow(x+1)(x+2)<0 \Rightarrow-2<x<-1
\end{array}
$So, $f(x)$ is decreasing, if $x \in(-2,-1)$.
\Rightarrow f^{\prime}(x)=6 x^2+18 x+12
$
For decreasing, $f^{\prime}(x)<0$
$
\begin{array}{ll}
\therefore & 6 x^2+18 x+12<0 \\
\Rightarrow & x^2+3 x+2<0 \Rightarrow(x+1)(x+2)<0 \Rightarrow-2<x<-1
\end{array}
$So, $f(x)$ is decreasing, if $x \in(-2,-1)$.
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