Find intervals in which the function given by $f\left(x\right)=\frac{3}{10}{x}^4-\frac{4}{5}{x}^3-3x^2+\frac{36}{5}+11$is (a) strictly increasing (b) strictly decreasing.

Solution: $f\left(x\right)=\frac{3}{10}{x}^4-\frac{4}{5}{x}^3-3x^2+\frac{36}{5}x+11$
$f^{\prime }\left(x\right)=\frac{6}{5}{x}^3-\frac{12}{5}{x}^2-6x+\frac{36}{5}$
If we put $f^{\prime }\left(x\right)=0$
$\frac{6}{5}{x}^3-\frac{12}{5}{x}^2-6x+\frac{36}{5}=0$
$\Rightarrow x^3-2x^2-5x+6=0$
$\Rightarrow \left(x-1\right)\left(x+2\right)\left(x-3\right)=0$
$\Rightarrow x=1,-2,3$
Now, we will draw these values of $x$ on number line.
Please refer to video to see the number line.
From the number line, we can see that,
For$x\in \left(-2,1\right)\cup \left(3,\infty \right),f\left(x\right)$ is increasing.
For $x\in \left(-\infty ,-2\right)\cup \left(1,3\right),f\left(x\right)$ is decreasing.