Question

The interval in which the function f is given by f(x) = x² e^-x is strictly increasing, is: ____________.

Options

• ${\displaystyle (-\infty ,0)}$

• (0, 2)

• ${\displaystyle (-\infty ,\infty )}$

• ${\displaystyle (2,\infty )}$

Answer 1

The interval in which the function f is given by f(x) = x² e^-x is strictly increasing, is: (0, 2).

Explanation:

Here,it is given that the function ${\displaystyle \text{f'}\left(\text{x}\right)=\frac{\text{d}}{\text{dx}}\left({\text{x}}^2{\text{e}}^{\text{-x}}\right)}$

${\displaystyle ={\text{x}}^2{\text{e}}^{\text{-x}}(-1)+2{\text{xe}}^{\text{-x}}}$

${\displaystyle \Rightarrow \text{f'}\left(\text{x}\right)={\text{e}}^{\text{-x}}(2\text{x}-{\text{x}}^2)>0}$

if ${\displaystyle (2\text{x}-{\text{x}}^2)>0}$

i.e. if x(x – 2 ) < 0 i.e. if 0 < x < 2. Hence f is strictly increasing on (0,2)