The largest interval, in which the function $f(x) = x^3 + 2x^2-1$ is increasing, is:

$\left(-∞,\frac{-4}{3},\right]∪[0,∞)$

The correct answer is Option (4) → $\left(-∞,\frac{-4}{3},\right]∪[0,∞)$

Given: $f(x)=x^{3}+2x^{2}-1$

Compute derivative: $f'(x)=3x^{2}+4x=x(3x+4)$

For $f(x)$ to be increasing: $f'(x)\ge0$

$x(3x+4)\ge0 \Rightarrow$ sign of $f'(x)$ changes at $x=0$ and $x=-\frac{4}{3}$.

Sign analysis:

• For $x<-\frac{4}{3}$ → both terms negative → $f'(x)>0$

• For $-\frac{4}{3}

• For $x>0$ → both terms positive → $f'(x)>0$

Hence $f(x)$ is increasing in intervals $(-\infty,-\frac{4}{3}]$ and $[0,\infty)$.

Required interval: $(-\infty,-\frac{4}{3}] \cup [0,\infty)$