Question

Find the interval in which the function $f\left(x\right)=2x^2-3x$ is
1) strictly increasing
2) strictly decreasing

Answer 1

Given that:

$f\left(x\right)=2x^2-3x$

Now, Differentiate w.r.t $x$

$\Rightarrow \frac{df\left(x\right)}{dx}=\frac{d(2x^2-3x)}{dx}=4x-3$

Now,

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

$\Rightarrow 4x-3=0$

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

Now,

$\left(i\right)$

strictly increasing, i.e., $f^{\prime }\left(x\right)>0,$

$x>\frac{3}{4}$

$x\in (\frac{3}{4},\infty )$

$\left(ii\right)$

strictly decreasing i.e., $f^{\prime }\left(x\right)<0$

$x<\frac{3}{4}$

$x\in (\infty ,\frac{3}{4})$