Question

Find the intervals in which the function f given by $f\left(x\right)=2x^2-3x$ is

a) strictly increasing

b) strictly decreasing

Answer 1

Given, $f\left(x\right)=2x^2-3x$
Differenting w.r.t x, $f^{\prime }\left(x\right)=4x-3$
On putting f'(x)=0, we get $4x-3=0\Rightarrow x=\frac{3}{4}$
The point $x=\frac{3}{4}$ divides the real line into two disjoint intervals namely

$(-\infty ,\frac{3}{4})$ and $(\frac{3}{4},\infty )$

$$\begin{array}{ccc}Intervals& Signof{f}^{\prime }\left(x\right)& Natureoff\left(x\right)\\ (-\infty ,\frac{3}{4})& (-)negative& Strictlydecreasing\\ & (\because x<\frac{3}{4})& \\ (\frac{3}{4},\infty )& (+)positive& Strictlyincreasing\\ & (\because x>\frac{3}{4})& \end{array}$$
Hence, f(x) is strictly increasing $(\frac{3}{4},\infty )$ and strictly decreasing on $(-\infty ,\frac{3}{4})$.