In which of these intervals is the function f(x) = 3x^2 - 4x strictly…

The correct answer is Option (1) → $(-\infty, 0)$ ##

Given,

$f(x) = 3x^2 - 4x + 1$

$∴f'(x) = \frac{d}{dx}(3x^2 - 4x + 1)$

$= 3 \cdot 2x - 4$

$= 6x - 4$

For decreasing, $f'(x) < 0$ or $6x - 4 < 0$

i.e., $x < \frac{4}{6} = \frac{2}{3}$

$∴x \in (-\infty, \frac{2}{3})$