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

Solution:

Given:- Function $f(x)=x^{3}-6 x^{2}+9 x+15$

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

(i) If $f^{\prime}(x)>0$ for $a l l_{x} \in(a, b)$, 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,

$f(x)=x^{3}-6 x^{2}+9 x+15$

$\Rightarrow f(x)=\frac{d}{d x}\left(x^{3}-6 x^{2}+9 x+15\right)$

$\Rightarrow f^{\prime}(x)=3 x^{2}-12 x+9$

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

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

$\Rightarrow 3 x^{2}-12 x+9=0$

$\Rightarrow 3\left(x^{2}-4 x+3\right)=0$

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

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

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

$\Rightarrow x=1,3$

clearly, $f^{\prime}(x)>0$ if $x<1$ and $x>3$

and $f^{\prime}(x)<0$ if $1

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

and $f(x)$ is decreasing on interval $x \in(1,3)$