Question

Let $f: R \rightarrow R$ be defined
as $ f(x)=\begin{cases}-\frac{4}{3} x^{3}+2 x^{2}+3 x, & x>0 \\ 3 x e^{x} & , x \leq 0\end{cases}$. Then $f$ is increasing function in the interval

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

Answer 1

Answer: (C)

$f^{\prime}(x) =\begin{cases}-4 x^{3}+4 x+3 & x>0 \\ 3 e^{x}(1+x) & x \leq 0\end{cases}$

For $x>0, f^{\prime}(x)=-4 x^{2}+4 x+3$
$f(x)$ is increasing in $\left(-\frac{1}{2}, \frac{3}{2}\right)$
For $x \leq 0, f^{\prime}(x)=3 e^{x}(1+x)$
$f^{\prime}(x) > 0 \forall x \in(-1,0)$
$\Rightarrow f(x)$ is increasing in $(-1,0)$
So, in complete domain, $f(x)$ is increasing in $\left(-1, \frac{3}{2}\right)$