Question

Find the intervals in which the function $f$ given by $f(x) = 2x^2 − 3x$ is (a) strictly increasing (b) strictly decreasing

Answer 1

The given function is $f(x) = 2x^2 − 3x$.
$f'(x)=4x-3$
$f'(x)=0\implies x=\frac{3}{4}$
Now, the point divides the real line into two disjoint intervals i.e., $(-∞,\frac{3}{4})$and$(\frac{3}{4},∞)$In interval $(-∞,-2)$ and $(3,∞),f'(x)$ is positive while in interval $(-2,3), f'(x)$ is negative.
Hence, the given function $(f)$ is strictly increasing in intervals $(-∞,-2)$ and $(3,∞)$, while function $(f)$ is strictly decreasing in interval $(−2, 3)$.