Find the intervals in which the following functions are increasing or decreasing. (i) $f(x)=2 x^{3}-15 x^{2}+36 x+1$ (ii) $f(x)=2 x^{3}+9 x^{2}+12 x+20$

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Find the intervals in which the following functions are increasing or decreasing.
(i) $f(x)=2 x^{3}-15 x^{2}+36 x+1$
(ii) $f(x)=2 x^{3}+9 x^{2}+12 x+20$

Find the intervals in which the following functions are increasing or decreasing.
(i) $f(x)=2 x^{3}-15 x^{2}+36 x+1$
(ii) $f(x)=2 x^{3}+9 x^{2}+12 x+20$

Solution:

(i) Given that $f(x)=2 x^{3}-15 x^{2}+36 x+1$
Now with respect to $x$ differentiating above equation, we obtain
$\begin{array}{l} f(x)=\frac{d}{d x}\left(2 x^{3}-15 x^{2}+36 x+1\right) \\ \Rightarrow f^{\prime}(x)=6 x^{2}-30 x+36 \end{array}$
For $f(x)$ we have to find critical point, we must have
$\begin{array}{l} \Rightarrow f^{\prime}(x)=0 \\ \Rightarrow 6 x^{2}-30 x+36=0 \\ \Rightarrow 6\left(x^{2}-5 x+6\right)=0 \\ \Rightarrow 6\left(x^{2}-3 x-2 x+6\right)=0 \\ \Rightarrow x^{2}-3 x-2 x+6=0 \\ \Rightarrow(x-3)(x-2)=0 \\ \Rightarrow x=3,2 \end{array}$
So, $f^{\prime}(x)>0$ if $x<2$ and $x>3$ and $f^{\prime}(x)<0$ if $2<x<3$
Therefore, $f(x)$ increases on $(-\infty, 2) \cup(3, \infty)$ and $f(x)$ is decreasing on interval $x \in(2,3)$

(ii) Given that $f(x)=2 x^{3}+9 x^{2}+12 x+20$
By differentiating above equation we obtain
$\begin{array}{l} \Rightarrow \\ f(x)=\frac{d}{d x}\left(2 x^{3}+9 x^{2}+12 x+20\right) \\ \Rightarrow f^{\prime}(x)=6 x^{2}+18 x+12 \end{array}$
For $f(x)$ we have to find critical point we must have
$\begin{array}{l} \Rightarrow f^{\prime}(x)=0 \\ \Rightarrow 6 x^{2}+18 x+12=0 \\ \Rightarrow 6\left(x^{2}+3 x+2\right)=0 \\ \Rightarrow 6\left(x^{2}+2 x+x+2\right)=0 \\ \Rightarrow x^{2}+2 x+x+2=0 \\ \Rightarrow(x+2)(x+1)=0 \\ \Rightarrow x=-1,-2 \end{array}$
So, $f^{\prime}(x)>0$ if $-2<x<-1$ and $f^{\prime}(x)<0$ if $x<-1$ and $x>-2$
Therefore, $f(x)$ increases on $x \in(-2,-1)$ and $f(x)$ is decreasing on interval $(-\infty,-2) \cup(-2, \infty)$