Question
The real function $f(x)=2 x^3-3 x^2-36 x+7$ is
✓
Answer
We have $, f(x)=2 x^3-3 x^2-36 x+7$
$\Rightarrow f^{\prime}(x)=6 x^2-6 x-36=6\left(x^2-x-6\right)$
$=6(x-3)(x+2)$
$f^{\prime}(x)=0 $
$\Rightarrow(x-3)(x+2)=0, x=-2,3$
$\therefore f(x)$ is strictly increasing in $(-\infty,-2) \cup(3, \infty)$ and strictly
$\Rightarrow f^{\prime}(x)=6 x^2-6 x-36=6\left(x^2-x-6\right)$
$=6(x-3)(x+2)$
$f^{\prime}(x)=0 $
$\Rightarrow(x-3)(x+2)=0, x=-2,3$
$\therefore f(x)$ is strictly increasing in $(-\infty,-2) \cup(3, \infty)$ and strictly
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