Determine the intervals in which the function f(x) = x^4-(4x^3)/3 are…

The correct answer is Option (3) → Strictly increasing on $[1,∞)$ and strictly decreasing on $(−∞,1]$.

Given $f(x) = x^4-\frac{4x^3}{3},D_f=R$

Diff. w.r.t. x, we get $f'(x) = 4x^3-\frac{4}{3}.3x^2 = 4x^2(x-1)$.

Now $f'(x) > 0$ iff $4x^2(x-1)>0⇒x-1>0$  $(∵x^2>0,x≠0)$

$⇒ x>1⇒x∈ (1,∞)$

⇒ f is strictly increasing in [1, ∞).

And $f'(x) < 0$ iff $4x^2(x-1) <0⇒x-1<0$

$⇒ x <1⇒x∈ (-∞, 1)$.

Note that $f'(x) < 0$ for all $x ∈ (-∞, 1)$ except $x = 0$ where $f'(x) = 0$.

Hence, f is strictly decreasing in (-∞, 1].