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The function $f(x)=5+36 x+3 x^{2}-2 x^{3}$ is increasing in the interval
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Verified Answer
The correct answer is:
$(-2,3)$
Given, $f(x)=5+36 x+3 x^{2}-2 x^{3}$
$\Rightarrow f^{\prime}(x)=36+6 x-6 x^{2}$
$=-6\left(x^{2}-x-6\right)$
$=-6\left(x^{2}-3 x+2 x-6\right)$
$=-6(x-3)(x+2)$
For increasing function, $f^{\prime}(x)>0$
$\Rightarrow \quad(x-3)(x+2) < 0$
$\Rightarrow \quad x < 3$ and $x>-2$
$\Rightarrow \quad-2 < x < 3$
Hence, function $f(x)$ is increasing $\operatorname{in}(-2,3)$ interval.
$\Rightarrow f^{\prime}(x)=36+6 x-6 x^{2}$
$=-6\left(x^{2}-x-6\right)$
$=-6\left(x^{2}-3 x+2 x-6\right)$
$=-6(x-3)(x+2)$
For increasing function, $f^{\prime}(x)>0$
$\Rightarrow \quad(x-3)(x+2) < 0$
$\Rightarrow \quad x < 3$ and $x>-2$
$\Rightarrow \quad-2 < x < 3$
Hence, function $f(x)$ is increasing $\operatorname{in}(-2,3)$ interval.
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