Let $f(x)=\left\{\begin{array}{l}x e^{a x} \quad, x \leq 0 \\ x+a x^2-x^3, x>0\end{array}\right.$, where a is a positive constant. Then, the interval in which f'(x) is increasing, is
Answer & explanation
Correct answer: option 3
We have,
$f(x)= \begin{cases}x e^{a x} & , x \leq 0 \\ x+a x^2-x^3 & , x>0\end{cases}$
Clearly, f(x) is continuous and differentiable at x = 0
∴ $f'(x)= \begin{cases}(1+a x) e^{a x} & , x \leq 0 \\ 1+2 a x-3 x^2 & , x>0\end{cases}$
f'(x) is also continuous and differentiable at x = 0
∴ $f''(x)= \begin{cases}(2+a x) a e^{a x} & , x \leq 0 \\ 2 a-6 x & , x>0\end{cases}$
For f'(x) to be increasing, we must have f''(x) > 0. Now two cases arise:
CASE I When x ≤ 0 :
In this case, we have
f''(x) > 0
$\Rightarrow (2+a x) a e^{a x}>0$
$\Rightarrow 2+a x>0$ [∵ a > 0 and $e^{a x}>0$]
$\Rightarrow x>-\frac{2}{a}$
∴ x ∈ (-2/a, 0] ........(i)
CASE II When x > 0 :
In this case, we have
f''(x) > 0
$\Rightarrow 2 a-6 x>0 \Rightarrow 2 a>6 x \Rightarrow x<\frac{a}{3}$
∴ $x \in(0, a / 3)$ .......(ii)
From (i) and (ii), we find that f'(x) is increasing for
$x \in(-2 / a, a / 3)$