Draw the graph of `f(x) = x^(2)e^(-|x|)`
i) Find the point of maxima/minima.
ii) Find the asymptote is any.
iii) Find the range of the function.
iv) Find the number of roots of the equation `f(x)=1`

Explanation

we have `f(x) = x^(2)e^(-|x|)`
Clearly, `f(x)=f(-x)`.
So `f(x)` is an even function.
Hence the graph is symmetrical about the y-axis.
For `x ge0, f(x) = x^(2)e^(-x)`
Now `f^(')(x) = -x^(2)e^(-x) + 2xe^(-x) =xe^(-x)(2-x)`
`f^(')(x) = -x^(2)e^(-x)+2xe^(-x) = xe^(-x)(2-x)`
`f^(')(x)=0 rArr x=0` or `x=2`
`f(0) =0` and `underset(x to infty)"lim"x^(2)/e^(x) = underset(x to infty)"lim"(2x)/e^(x)=underset(x to infty)2/e^(x)=0`
So x=2 is the pont of maxima.
Hence the graph of `f(x)` is as shown in the following figure.

Since, `f(x)` is differentiable at `x=0`,the graph touches the x-axis at x=0
The x-axis is also asymptote to the curve.
Also since `f(x)` is an even function, `x=-2` is also the point of maxima.
Range of the function is `[0,f(2)]` or `[0, 4//e^(2)]`
Clearly, `f(x) ne 1`, hence `f(x)=1` has no roots.