question_answer

Find the interval in which the function f given by \[f(x)=2{{x}^{2}}-3x\] is strictly increasing.

Answer:

We have, \[f\,(x)=2{{x}^{2}}-3x\] On differentiating w.r.t. x, we get \[f'(x)=4x-3\] On putting\[f'(x)=0,\] we get \[4x-3=0\Rightarrow 4x=3\Rightarrow x=\frac{3}{4}\] Here, point\[x=\frac{3}{4}\]divide the real line into 2 disjoint intervals i.e.,\[\left( -\,\infty ,\,\,\frac{3}{4} \right)\]and \[\left( \frac{3}{4},\,\,\infty  \right)\] \[\therefore \]For \[x>\frac{3}{4},\]value of \[f'(x)>0\] and for \[x<\frac{3}{4},\]value of \[f'(x)<0\] Hence, \[f\,(x)\]is strictly increasing in the interval\[\left( \frac{3}{4},\,\,\infty  \right)\].