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

Solution:

Given:- Function $\mathrm{f}(\mathrm{x})=\frac{1}{4} \mathrm{x}^{4}+\frac{2}{3} \mathrm{x}^{3}-\frac{5}{2} \mathrm{x}^{2}-6 \mathrm{x}+7$

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,

$f(x)=\frac{1}{4} x^{4}+\frac{2}{3} x^{3}-\frac{5}{2} x^{2}-6 x+7$

$\Rightarrow f(x)=\frac{d}{d x}\left(\frac{1}{4} x^{4}+\frac{2}{3} x^{3}-\frac{5}{2} x^{2}-6 x+7\right)$

$\Rightarrow f^{\prime}(x)=x^{3}+2 x^{2}-5 x-6$

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

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

$\Rightarrow x^{3}+2 x^{2}-5 x-6=0$

$\Rightarrow(x+1)(x-2)(x+3)=0$

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

clearly, $f^{\prime}(x)>0$ if $-32$

and $f^{\prime}(x)<0$ if $x<-3$ and $-3

Thus, $f(x)$ increases on $(-3,-1) \cup(2, \infty)$

and $f(x)$ is decreasing on interval $(\infty,-3) \cup(-1,2)$