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

Solution:

Given:- Function $f(x)=x^{3}-12 x^{2}+36 x+17$

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,

$f(x)=x^{3}-12 x^{2}+36 x+17$

$\Rightarrow f(x)=\frac{d}{d x}\left(x^{3}-12 x^{2}+36 x+17\right)$

$\Rightarrow f^{\prime}(x)=3 x^{2}-24 x+36$

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

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

$\Rightarrow 3 x^{2}-24 x+36=0$

$\Rightarrow 3\left(x^{2}-8 x+12\right)=0$

$\Rightarrow 3\left(x^{2}-6 x-2 x+12\right)=0$

$\Rightarrow x^{2}-6 x-2 x+12=0$

$\Rightarrow(x-6)(x-2)=0$

$\Rightarrow x=2,6$

clearly, $f^{\prime}(x)>0$ if $x<2$ and $x>6$

and $f^{\prime}(x)<0$ if $2

Thus, $f(x)$ increases on $(-\infty, 2) \cup(6, \infty)$

and $f(x)$ is decreasing on interval $x \in(2,6)$