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

Solution:

Given:- Function $f(x)=(x-1)(x-2)^{2}$

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,

$f(x)=(x-1)(x-2)^{2}$

$\Rightarrow \mathrm{f}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left((\mathrm{x}-1)(\mathrm{x}-2)^{2}\right)$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=(\mathrm{x}-2)^{2}+2(\mathrm{x}-2)(\mathrm{x}-1)$

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

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=(\mathrm{x}-2)(3 \mathrm{x}-4)$

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

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

$\Rightarrow(x-2)(3 x-4)=0$

$\Rightarrow x=2, \frac{4}{3}$

clearly, $f^{\prime}(x)>0$ if $x<\frac{4}{3}$ and $x>2$

and $f^{\prime}(x)<0$ if $\frac{4}{3}

Thus, $f(x)$ increases on $\left(-\infty, \frac{4}{3}\right) \cup(2, \infty)$

and $f(x)$ is decreasing on interval $x \in\left(\frac{4}{3}, 2\right)$