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If $f(x)=2 x^3-15 x^2-144 x-7$, then $f(x)$ is strictly decreasing in
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Verified Answer
The correct answer is:
$(-3,8)$
$$
\begin{aligned}
& \mathrm{f}(\mathrm{x})=2 \mathrm{x}^3-15 \mathrm{x}^2-144 \mathrm{x}-7 \\
& \mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}^2-30 \mathrm{x}-144 \\
& \text { When } \mathrm{f}^{\prime}(\mathrm{x}) < 0, \text { we get } \mathrm{x}^2-5 \mathrm{x}-24 < 0 \\
& \therefore(\mathrm{x}-8)(\mathrm{x}+3) < 0 \Rightarrow-3 < \mathrm{x} < 8
\end{aligned}
$$
\begin{aligned}
& \mathrm{f}(\mathrm{x})=2 \mathrm{x}^3-15 \mathrm{x}^2-144 \mathrm{x}-7 \\
& \mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}^2-30 \mathrm{x}-144 \\
& \text { When } \mathrm{f}^{\prime}(\mathrm{x}) < 0, \text { we get } \mathrm{x}^2-5 \mathrm{x}-24 < 0 \\
& \therefore(\mathrm{x}-8)(\mathrm{x}+3) < 0 \Rightarrow-3 < \mathrm{x} < 8
\end{aligned}
$$
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