Let $f(x)=\left\{\begin{array}{cc}x e^{a x} & x \leq 0 \ x+a x^{2}-x^{3}, & x0\end{array} ight.$

Question

Let $f(x)=\left\{\begin{array}{cc}x e^{a x} & x \leq 0 \ x+a x^{2}-x^{3}, & x>0\end{array}ight.$ Where $a$ is a positive constant. The interval in which $f^{\prime}(x)$ is increasing is

Professor PC Thomas Classes · Answer

Given that,\ $f(x)=\left\{\begin{array}{cc}x e^{a x}, & x \leq 0 \ x+a x^{2}-x^{3}, & x>0\end{array}ight.$\ Differentiating both sides, we have\ $f^{\prime}(x)=\left\{\begin{array}{cc}a x e^{a x}+e^{a x}, & x \leq 0 \ 1+2 a x-3 x^{2}, & x>0\end{array}ight.$\ Again differentiating both sides, we have\ $f^{\prime \prime}(x)=\left\{\begin{array}{cc}2 a e^{a x}+a^{2} x e^{a x} ; & x \leq 0 \ 2 a-6 x ; & x>0\end{array}ight.$\ For ciritical points, we put $f^{\prime \prime}(x)=0$\ $\Rightarrow x=-\frac{2}{a}$, if $x \leq 0$\ $x=\frac{a}{3}$, if $x>0$\ It is clear from number line that\ $f^{\prime \prime}(x)$ is +ve on $\left(-\frac{2}{a}, \frac{a}{3}ight) \Rightarrow f^{\prime}(x)$ increases on $\left(-\frac{2}{a}, \frac{a}{3}ight)$.\