Question

Let $f\left(x\right)=\{\begin{array}{cc}x{e}^{ax}& x\le 0\\ x+a{x}^2-x^3& x>0\end{array}$ , where $a$ is positive constant. Find the interval in which $f\left(x\right)$ is increasing.

• A. $\left[\frac{-2}{a},\frac{a}{3}\right]$
• B. $\left[\frac{2}{a},\frac{-a}{3}\right]$
• C. $\left[\frac{-2}{a},\frac{a}{6}\right]$
• D. $\left[\frac{-3}{a},\frac{a}{2}\right]$

Answer 1

The correct option is A $\left[\frac{-2}{a},\frac{a}{3}\right]$

At $x=0,L.H.L=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}{xe}^{ax}=0$

and $R.H.L=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}(x+{ax}^2-x^3)=0$

Therefore, $L.H.L=R.H.L=0=f\left(0\right)$

So, $f\left(x\right)$ is continuous at $x=0.$

Also, $f^{\prime }\left(x\right)=\{\begin{array}{c}\begin{array}{cc}1.e^{ax}+ax{e}^{ax},& x<0\end{array}\\ \begin{array}{cc}1+2ax-3x^2,& x>0\end{array}\end{array}$

and $L{f}^{\prime }\left(0\right)=\underset{x\to 0^{-}}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0}$ $=\underset{x\to 0^{-}}{lim}\frac{{xe}^{ax}-0}{x}=\underset{x\to 0^{-}}{lim}{e}^{ax}=e^0=1$

and $R{f}^{\prime }\left(0\right)=\underset{x\to 0^{+}}{lim}\frac{f\left(x\right)-f\left(0\right)}{x+0}$ $=\underset{x\to 0^{+}}{lim}\frac{x+{ax}^2-x^3-0}{x}$ $=\underset{x\to 0^{+}}{lim}1+ax-x^2=1$

Therefore, $L{f}^{\prime }\left(0\right)=R{f}^{\prime }\left(0\right)=1\Rightarrow f^{\prime }\left(0\right)=1$

Hence, $f^{\prime }\left(x\right)=\{\begin{array}{c}\begin{array}{cc}(ax+1)e^{ax},& x<0\end{array}\\ \begin{array}{cc}1,& x=0\end{array}\\ \begin{array}{cc}1+2ax-{3x}^2,& x>0\end{array}\end{array}$

Now, we can say without solving that, $f^{\prime }\left(x\right)$bis continuous at $x=0$ and hence on $R.$

We have,

$f^{\prime }\left(x\right)=\{\begin{array}{c}\begin{array}{cc}{ae}^{ax}+a(ax+1)e^{ax},& x<0\end{array}\\ \begin{array}{cc}2a-6x,& x>0\end{array}\end{array}$

and $L{f}^{\prime \prime }\left(0\right)=\underset{x\to 0^{-}}{lim}\frac{f^{\prime }\left(x\right)-f^{\prime }\left(0\right)}{x-0}=\underset{x\to 0^{-}}{lim}\frac{(ax-1)e^{ax}-1}{x}$ $=\underset{x\to 0^{-}}{lim}[{ae}^{ax}+\frac{e^{ax}-1}{x}]$

$=\underset{x\to 0^{-}}{lim}{ae}^{ax}+a.\underset{x\to 0^{-}}{lim}\frac{e^{ax}-1}{ax}$ $={ae}^0+a\left(1\right)=2a$

and $R{f}^{\prime \prime }\left(0\right)=\underset{x\to 0^{+}}{lim}\frac{f^{\prime }\left(x\right)-f^{\prime }\left(0\right)}{x+0}$ $=\underset{x\to 0^{+}}{lim}\frac{(1+2ax-{3x}^2)-1}{x}$ $=\underset{x\to 0^{+}}{lim}\frac{2ax-{3x}^2}{x}=\underset{x\to 0^{+}}{lim}2a-3x=2a$

Therefore $L{f}^{\prime \prime }\left(0\right)=R{f}^{\prime \prime }\left(0\right)=2a$

Hence, $f^{\prime \prime }\left(x\right)=\{\begin{array}{c}\begin{array}{cc}a(ax+2)e^{ax},& x<0\end{array}\\ \begin{array}{cc}2a,& x=0\end{array}\\ \begin{array}{cc}2a-6x,& x>0\end{array}\end{array}$

Now, for $x<0,f^{\prime \prime }\left(x\right)>0$, if $ax+2>0$

$\Rightarrow$ for $x<0,f^{\prime \prime }\left(x\right)>0,$, if $x>-\frac{2}{a}$

$\Rightarrow f^{\prime \prime }\left(x\right)>0$ if $-\frac{2}{a}<x<0$

And for $x>0,f^{\prime \prime }\left(x\right)>0,$ if $2a-6x>0$

$\Rightarrow$ for $x>0,f^{\prime \prime }\left(x\right)>0,$ if $x<\frac{a}{3}$

$f\left(x\right)$ increases on $[-\frac{2}{a},0]$ and on $[0,\frac{a}{3}]$

Hence, $f\left(x\right)$ increases on $[-\frac{2}{a},\frac{a}{3}].$