Practicing Success
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 |
$(0, a / 3)$ $(-2 / a, 0)$ $(-2 / a, a / 3)$ none of these |
$(-2 / a, a / 3)$ |
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 $x \in(-2 / a, a / 3)$ |