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

Solution:

Given:- Function $f(x)=x^{2}+2 x-5$

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,

$f(x)=x^{2}+2 x-5$

$\Rightarrow f(x)=\frac{d}{d x}\left(x^{2}+2 x-5\right)$

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

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

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

$\Rightarrow 2 x+2>0$

$\Rightarrow 2 x<-2$

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

$\Rightarrow x<-1$

$\Rightarrow x \in(-\infty,-1)$

Thus $f(x)$ is increasing on interval $(-\infty,-1)$

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

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

$\Rightarrow 2 x+2<0$

$\Rightarrow 2 x>-2$

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

$\Rightarrow x>-1$

$\Rightarrow x \in(-1, \infty)$

Thus $f(x)$ is decreasing on interval $x \in(-1, \infty)$