Question

Let $f\left(x\right)=x^2.e^{-x^2}$ then

• A. f(x) has local maxima at x = -1 and x = 1
• B. f(x) has local minima at x = 0
• C. f(x) is strictly decreasing on x $ϵ$ R
• D. Range of f(x) is $[0,\frac{1}{e}]$

Answer 1

Answer: (A), (B), (D)

The correct options are
A f(x) has local maxima at x = -1 and x = 1
B f(x) has local minima at x = 0
D Range of f(x) is $[0,\frac{1}{e}]$

$f\left(x\right)=x^2.e^{-x^2}$
$f^{\prime }\left(x\right)=2x.e^{-x^2}+x^2.e^{-x^2}(-2x)$
$=2x{e}^{-x^2}[1-x^2]$

f(x) has local maxima at x = -1 and 1
f(x) has local minima at x = 0
Now; f(0) = 0
$f\left(1\right)=\frac{1}{e}$ and as $x\to \infty ,f\left(x\right)\to 0$
So, Range of f(x) is $(0,\frac{1}{e})$.