# Find the intervals in which the function $f$ given by $f (x) = 2x^2 - 3x$ is $(a) \;strictly\; increasing$

$\begin{array}{1 1} increasing \;in \;(-\infty,\frac{-3}{4}) decreasing\; in \;(\frac{-3}{4},\infty) \\ increasing\; in \;(-\infty,\frac{-3}{4}) decreasing \;in \;(\frac{3}{4},\infty) \\increasing \;in \;(-\infty,\frac{3}{4}) decreasing\; in\; (\frac{3}{4},\infty) \\ decreasing \;in\; (-\infty,\frac{3}{4}) increasing \;in \;(\frac{3}{4},\infty)\end{array}$

Toolbox:
• A function $f(x)$ is said to be a strictly increasing function on $(a,b)$ if $x_1 < x_2\Rightarrow f(x_1) < f(x_2)$ for all $x_1,x_2\in (a,b)$
• If $x_1 < x_2\Rightarrow f(x_1) > f(x_2)$ for all $x_1,x_2\in (a,b)$ then $f(x)$ is said to be strictly decreasing on $(a,b)$
• A function $f(x)$ is said to be increasing on $[a,b]$ if it is increasing (decreasing) on $(a,b)$ and it is increasing (decreasing) at $x=a$ and $x=b$.
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly increasing on $(a,b)$ is that $f'(x) > 0$ for all $x\in (a,b)$
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly decreasing on $(a,b)$ is that $f'(x) < 0$ for all $x\in (a,b)$
Step 1:
Given :$f(x)=2x^2-3x$
Differentiating w.r.t $x$ we get,
$f'(x)=4x-3$
$f'(x)=0$
$\Rightarrow 4x-3=0$
$x=\large\frac{3}{4}$
Step 2:
This point $x=\large\frac{3}{4}$ divides the real number line into two intervals (i.e) $(\infty,\large\frac{3}{4})$ and $(\large\frac{3}{4},$$\infty) Consider the interval (\large\frac{3}{4},$$\infty)$
Here $f'(x)$ is positive.
Hence it is strictly increasing in $(\large\frac{3}{4},$$\infty)$