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The range of values of $x$ for which $f(x)=x^3+6 x^2-36 x+7$ is increasing in
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Verified Answer
The correct answer is:
$(-\infty,-6) \cup(2, \infty)$
$\begin{aligned}
\mathrm{f}(x) & =x^3+6 x^2-36 x+7 \\
\mathrm{f}^{\prime}(x) & =3 x^2+12 x-36 \\
& =3\left(x^2+4 x-12\right)
\end{aligned}$
For $\mathrm{f}(x)$ to be increasing,
$\begin{aligned}
& \mathrm{f}^{\prime}(x)>0 \\
& \Rightarrow 3\left(x^2+4 x-12\right)>0 \\
& \Rightarrow x^2+4 x-12>0 \\
& \Rightarrow(x+6)(x-2)>0 \\
& \Rightarrow x \in(-\infty,-6) \cup(2, \infty)
\end{aligned}$
\mathrm{f}(x) & =x^3+6 x^2-36 x+7 \\
\mathrm{f}^{\prime}(x) & =3 x^2+12 x-36 \\
& =3\left(x^2+4 x-12\right)
\end{aligned}$
For $\mathrm{f}(x)$ to be increasing,
$\begin{aligned}
& \mathrm{f}^{\prime}(x)>0 \\
& \Rightarrow 3\left(x^2+4 x-12\right)>0 \\
& \Rightarrow x^2+4 x-12>0 \\
& \Rightarrow(x+6)(x-2)>0 \\
& \Rightarrow x \in(-\infty,-6) \cup(2, \infty)
\end{aligned}$
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