Question

Let $f:R\to R$ be defined as
$f\left(x\right)=\{\begin{array}{cc}-\frac{4}{3}{x}^3+2x^2+3x,& x>0\\ 3x{e}^x,& x\le 0\end{array}$
Then $f$ is increasing function in the interval:

• A. $(0,2)$
• B. $(-\frac{1}{2},2)$
• C. $(-3,-1)$
• D. $(-1,\frac{3}{2})$

Answer 1

The correct option is D $(-1,\frac{3}{2})$

Given:$f\left(x\right)=\{\begin{array}{cc}-\frac{4}{3}{x}^3+2x^2+3x,& x>0\\ 3x{e}^x,& x\le 0\end{array}$
$f^{\prime }\left(x\right)=\{\begin{array}{ccc}-4x^2+4x+3& ,& x>0\\ 3e^x(x+1)& ,& x\le 0\end{array}$
Here $f\left(x\right)$ is differentiable at $x=0$
$\therefore f^{\prime }\left(x\right)=\{\begin{array}{ccc}4-(2x-1{)}^2& ,& x>0\\ 3e^x(x+1)& ,& x\le 0\end{array}$
Here $f^{\prime }\left(x\right)>0$ when $x\in (-1,\frac{3}{2})$
$\therefore f\left(x\right)$ is increasing in $(-1,\frac{3}{2})$