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

Solution:

Given:- Function $f(x)=\frac{3}{2} x^{4}-4 x^{3}-45 x^{2}+51$

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

(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,

$f(x)=\frac{3}{2} x^{4}-4 x^{3}-45 x^{2}+51$

$\Rightarrow \mathrm{f}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{3}{2} \mathrm{x}^{4}-4 \mathrm{x}^{3}-45 \mathrm{x}^{2}+51\right)$

$\Rightarrow f^{\prime}(x)=6 x^{3}-12 x^{2}-90 x$

$\Rightarrow f^{\prime}(x)=6 x\left(x^{2}-2 x-15\right)$

$\Rightarrow f^{\prime}(x)=6 x\left(x^{2}-5 x+3 x-15\right)$

$\Rightarrow f^{\prime}(x)=6 x(x-5)(x+3)$

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

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

$\Rightarrow 6 x(x-5)(x+3)>0$

$\Rightarrow x(x-5)(x+3)>0$

$\Rightarrow-3

$\Rightarrow x \in(-3,0) \cup(5, \infty)$

Thus $f(x)$ is increasing on interval $(-3,0) \cup(5, \infty)$

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

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

$\Rightarrow 6 x(x-5)(x+3)>0$

$\Rightarrow x(x-5)(x+3)>0$

$\Rightarrow-\infty

$\Rightarrow x \in(-\infty,-3) \cup(0,5)$

Thus $f(x)$ is decreasing on interval $(-\infty,-3) \cup(0,5)$