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

The given function is $f\left(x\right)=2x^2-3x$.
$\Rightarrow f^{\prime }\left(x\right)=4x-3$
$\therefore f^{\prime }\left(x\right)=0\Rightarrow x=\frac{3}{4}$
Now, the point $\frac{3}{4}$ divides the real line into two disjoint intervals i.e., $(-\infty ,\frac{3}{4})$ and $(\frac{3}{4},-\infty )$

In interval $(-\infty ,\frac{3}{4}),f^{\prime }\left(x\right)=4x-3<0.$

Hence, the given function $\left(f\right)$ is strictly decreasing in interval $(-\infty ,\frac{3}{4})$ In interval $(\frac{3}{4},\infty ),f^{\prime }\left(x\right)=4x-3>0.$
Hence, the given function $\left(f\right)$ is strictly increasing in interval $(\frac{3}{4},\infty )$.