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

Question:

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

$f(x)=8+36 x+3 x^{2}-2 x^{3}$

Solution:

Given:- Function $f(x)=8+36 x+3 x^{2}-2 x^{3}$

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,

$f(x)=8+36 x+3 x^{2}-2 x^{3}$

$\Rightarrow f(x)=\frac{d}{d x}\left(8+36 x+3 x^{2}-2 x^{3}\right)$

$\Rightarrow f^{\prime}(x)=36+6 x-6 x^{2}$

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

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

$\Rightarrow 36+6 x-6 x^{2}=0$

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

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

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

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

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

$\Rightarrow x=3,-2$

clearly, $f^{\prime}(x)>0$ if $-2 and$f^{\prime}(x)<0$if$x<-2$and$x>3$Thus,$f(x)$increases on$x \in(-2,3)$and$f(x)$is decreasing on interval$(-\infty,-2) \cup(3, \infty)\$

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