JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Application of Derivatives

For $x>0$: $f^{\prime }(x)=-4x^2+4x+3$. Setting $f^{\prime }(x)>0$ gives $-4x^2+4x+3>0$, or $4x^2-4x-3<0$. Solving $4x^2-4x-3=0$ using the quadratic formula: $x=\frac{4\pm \sqrt{16+48}}{8}=\frac{4\pm 8}{8}$, giving $x=\frac{3}{2}$ or $x=-\frac{1}{2}$. Since $x>0$, $f$ is increasing on $(0,\frac{3}{2})$.

For $x\le 0$: $f^{\prime }(x)=3e^x+3x{e}^x=3e^x(1+x)$. Since $e^x>0$, $f^{\prime }(x)>0$ when $1+x>0$, i.e., $x>-1$. So $f$ is increasing on $(-1,0]$.

Combining both regions, $f$ is increasing on $(-1,\frac{3}{2})$.