Question

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

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

Answer 1

Answer: (C)

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

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