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

Solution:

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

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,

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

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

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

For $f(x)$ to be increasing, we must have

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

$\Rightarrow-9-2 \mathrm{x}>0$

$\Rightarrow-2 \mathrm{x}>9$

$\Rightarrow \mathrm{x}<-\frac{9}{2}$

$\Rightarrow \mathrm{x}<-\frac{9}{2}$

$\Rightarrow \mathrm{x} \in\left(-\infty,-\frac{9}{2}\right)$

Thus $f(x)$ is increasing on interval $\left(-\infty,-\frac{9}{2}\right)$

Again, For $f(x)$ to be decreasing, we must have

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

$\Rightarrow-9-2 x<0$

$\Rightarrow-2 x<9$

$\Rightarrow x>-\frac{9}{2}$

$\Rightarrow x>-\frac{9}{2}$

$\Rightarrow x \in\left(-\frac{9}{2}, \infty\right)$

Thus $f(x)$ is decreasing on interval $x \in\left(-\frac{9}{2}, \infty\right)$