(i) $f(x)=3x-2$ where $f: \mathbb{R}\rightarrow \mathbb{R}$
# Method 1:
Let $x_1, \; x_2$ be any element of $\mathbb{R}$ such that:
$\hspace{1.5cm}x_1 > x_2$
$\hspace{1.3cm}3x_1 > 3x_2$
$\hspace{0.5cm}3x_{1}-2 > 3x_{2}-2$
$\hspace{1cm}f(x_1) > f(x_2)\hspace{0.5cm}\forall\hspace{0.5cm}x_{1},\;x_{2}\in\mathbb{R}$
So, $f(x)$ is strictly increasing on $\mathbb{R}.$
# Method 2:
$\hspace{1cm}f(x)=3x-2$
$\hspace{0.88cm}f'(x)=3$
$\hspace{0.17cm}$ As, $f'(x)>0\hspace{0.5cm}\hspace{0.5cm}\forall\hspace{0.5cm}x_{1},\;x_{2}\in\mathbb{R}$
So, $f(x)$ is strictly increasing on $\mathbb{R}.$
(ii) $f(x)=3-2x$ where $f: \mathbb{R}\rightarrow \mathbb{R}$
# Method 1:
Let $x_1, \; x_2$ be any element of $\mathbb{R}$ such that:
$\hspace{1.6cm}x_1 > x_2$
$\hspace{1cm}-2x_1 < -2x_2$
$\hspace{0.5cm}3-2x_{1} < 3-2x_{2}$
$\hspace{1cm}f(x_1) < f(x_2)\hspace{0.5cm}\forall\hspace{0.5cm}x_{1},\;x_{2}\in\mathbb{R}$
So, $f(x)$ is strictly decreasing on $\mathbb{R}.$
# Method 2:
$\hspace{1cm}f(x)=3-2x$
$\hspace{0.88cm}f'(x)=-2$
$\hspace{0.17cm}$ As, $f'(x)<0\hspace{0.5cm}\hspace{0.5cm}\forall\hspace{0.5cm}x_{1},\;x_{2}\in\mathbb{R}$
So, $f(x)$ is strictly decreasing on $\mathbb{R}.$