The interval on which the function f(x) = x^4 -(x^3)/3 is strictly de…

The correct answer is Option (3) → $(-∞,\frac{1}{4})$

Given: $f(x) = x^4 - \frac{x^3}{3}$

Compute first derivative: $f'(x) = 4x^3 - x^2$

$f'(x) = x^2(4x - 1)$

Now, find where $f'(x) < 0$:

$x^2(4x - 1) < 0$

$x^2 \ge 0$ always, so the sign of $f'(x)$ depends on $(4x - 1)$

$(4x - 1) < 0 \Rightarrow x < \frac{1}{4}$

Therefore, $f'(x) < 0$ only when $x < \frac{1}{4}$

The function is strictly decreasing on $(-\infty, \frac{1}{4})$