Find the intervals in which the following functions are increasing or decreasing.

Given:- Function $\mathrm{f}(\mathrm{x})=\frac{3}{10} \mathrm{x}^{4}-\frac{4}{5} \mathrm{x}^{3}-3 \mathrm{x}^{2}+\frac{36}{5} \mathrm{x}+11$

Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$.

(i) If $f^{\prime}(x)>0$ for all $x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$

(ii) If $f^{\prime}(x)<0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$

Algorithm:-

(i) Obtain the function and put it equal to $f(x)$

(ii) Find $f^{\prime}(x)$

(iii) Put $f^{\prime}(x)>0$ and solve this inequation.

For the value of $x$ obtained in (ii) $f(x)$ is increasing and for remaining points in its domain it is decreasing.

Here we have,

$\mathrm{f}(\mathrm{x})=\frac{3}{10} \mathrm{x}^{4}-\frac{4}{5} \mathrm{x}^{3}-3 \mathrm{x}^{2}+\frac{36}{5} \mathrm{x}+11$

$\Rightarrow \mathrm{f}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{3}{10} \mathrm{x}^{4}-\frac{4}{5} \mathrm{x}^{3}-3 \mathrm{x}^{2}+\frac{36}{5} \mathrm{x}+11\right)$

$\Rightarrow \mathrm{f}(\mathrm{x})=4 \times \frac{3}{10} \mathrm{x}^{3}-3 \times \frac{4}{5} \mathrm{x}^{2}-6 \mathrm{x}+\frac{36}{5}$

For $f(x)$ lets find critical point, we must have

$\Rightarrow f^{\prime}(x)=0$

$\Rightarrow \frac{12}{10} x^{3}-\frac{12}{5} x^{2}-6 x+\frac{36}{5}=0$

$\Rightarrow \frac{6}{5}(x-1)(x+2)(x-3)=0$

$\Rightarrow x=1,-2,3$

Now, lets check values of $f(x)$ between different ranges

Here points $x=1,-2,3$ divide the number line into disjoint intervals namely, $(-\infty,-2),(-2,1),(1,3)$ and $(3,$, $\infty)$

Lets consider interval $(-\infty,-2)$

In this case, we have $x-1<0, x+2<0$ and $x-3<0$

Therefore, $f^{\prime}(x)<0$ when $-\infty

Thus, $f(x)$ is strictly decreasing on interval $x \in(-\infty,-2)$

consider interval $(-2,1)$

In this case, we have $x-1<0, x+2>0$ and $x-3<0$

Therefore, $f^{\prime}(x)>0$ when $-2

Thus, $f(x)$ is strictly increases on interval $x \in(-2,1)$

Now, consider interval $(1,3)$

In this case, we have $x-1>0, x+2>0$ and $x-3<0$

Therefore, $f^{\prime}(x)<0$ when $1

Thus, $f(x)$ is strictly decreases on interval $x \in(1,3)$

finally, consider interval $(3, \infty)$

In this case, we have $x-1>0, x+2>0$ and $x-3>0$

Therefore, $f^{\prime}(x)>0$ when $x>3$

Thus, $f(x)$ is strictly increases on interval $x \in(3, \infty)$