Let $f(x)=\left\{\begin{matrix}xe^{ax}&;x≤0\\x+ax^2-x^3&;x>0\end{matrix}\right.$ where a is positive constant. The interval in which $f'(x)$ is increasing is
Answer & explanation
Correct answer: option 3
Given, $f(x)=\left\{\begin{matrix}xe^{ax}&;x≤0\\x+ax^2-x^3&;x>0\end{matrix}\right.$
Differentiating, both sides, we have
$f'(x)=\left\{\begin{matrix}axe^{ax}+e^{ax}&;x≤0\\1+2ax^2-3x^2&;x>0\end{matrix}\right.$
Again differentiating both sides, we have
$f''(x)=\left\{\begin{matrix}2ae^{ax}+a^2xe^{ax}&;x≤0\\2a-6x&;x>0\end{matrix}\right.$
Now $f''(x)= 0$ then in the interval the root is $x=-\frac{2}{a}$ and in interval when x > 0 root is $x=\frac{a}{3}$
Using sign scheme or number line rule, as shown in figure
Hence, $f'(x)$ decrease on $(-∞,-\frac{2}{a})∪(\frac{a}{3},∞)$ and increases on $(\frac{-2}{a},\frac{a}{3})$